Conic section

Class 12 Mathematics, unit 7, book pages 182–229

A conic section is the curve cut from a double cone by a plane: a circle, a parabola, an ellipse or a hyperbola. Each one is also a locus of points, defined by distances from a fixed point (focus) and a fixed line (directrix), and the ratio of these distances, the eccentricity $e$ , tells the four apart. This unit derives the standard equation of each conic, finds its elements (centre, vertices, foci, directrices, latus rectum, asymptotes), and works out tangents and normals. These curves describe real things: planetary and satellite orbits, satellite dishes and headlights, whispering galleries, bridges and cooling towers.

Learning outcomes (22)

Demonstrate conics and members of its family, i.e. circle, parabola, ellipse and hyperbola.
Derive and apply the equation of a circle in standard form, i.e. $(x - h)^{2} + (y - k)^{2} = r^{2}$ .
Find the equation of a circle passing through three non-collinear points; two points with its centre on a given line; two points when the equation of the tangent at one of these points is known; two points and touching a given line.
Find the condition when a line intersects the circle and when it touches the circle.
Find the equation of a tangent to a circle in slope form and of a normal to a circle at a point.
Find the length of the tangent to a circle from a given external point.
Derive and apply the standard equation of a parabola.
Sketch graphs of parabolas and find their elements.
Find the equation of a parabola with given elements such as focus and vertex, or focus and directrix.
Find the condition when a line is tangent to a parabola at a point and hence write the equation of a tangent line in slope form.
Find the equation of the tangent and normal to a parabola at a point.
Derive and apply the standard form of the equation of an ellipse and identify its elements.
Convert a given equation to the standard form of the equation of an ellipse, find its elements and draw the graph.
Find points of intersection of an ellipse with a line, including the condition of tangency.
Find the equation of a tangent to an ellipse in slope form.
Find the equation of a tangent and normal to an ellipse at a point.
Derive and apply the standard form of the equation of a hyperbola and identify its elements.
Find the equation of a hyperbola with given elements: transverse and conjugate axes with centre at the origin; two points; eccentricity, latera recta and transverse axis; focus, eccentricity and centre; focus, centre and directrix.
Find points of intersection of a hyperbola with a line, including the condition of tangency.
Find the equation of a tangent to a hyperbola in slope form.
Find the equation of a tangent and a normal to a hyperbola at a point.
Apply concepts of conics to real-life problems (suspension and reflection problems related to the parabola, satellite systems, elliptic motion of electrons around the nucleus, radio systems, flashlights, conics in architecture).

7.1Conics and its family members

A conic section (or simply a conic) is the curve you get when a flat plane cuts a right circular cone. Change the angle of the cut and you get four different curves: the circle, the ellipse, the parabola and the hyperbola. These four curves form one family. In this unit we find the equation of each member and study its tangents, normals and uses.

The right circular cone

Take a circle with centre A and a fixed point O on the line through A perpendicular to the plane of the circle. Join O to a point P on the circle and let the segment $\overline{O P}$ move once round the circle. The surface swept out is a right circular cone.

Vertex: The fixed point O of the cone.
Axis: The ray $O A$ from the vertex through the centre of the base circle. It is perpendicular to every radius such as $\overline{A P}$ , which is why the cone is called a right circular cone.
Generator (ruling): Any line segment such as $\overline{O P}$ that joins the vertex to a point of the circle.
Plane: A flat two-dimensional surface that extends without end in all directions.
Nappes: If every generator is extended through O, we get two cones joined vertex to vertex (a double cone). Each of the two cones is called a nappe.

A right circular cone: vertex O, axis

O A

, radius

\overline{A P}

and generator

\overline{O P}

Four ways to cut a cone

Circle: the plane is perpendicular to the axis and does not pass through the vertex.
Ellipse: the plane is tilted a little (not perpendicular to the axis), cuts every generator of one nappe and does not pass through the vertex.
Parabola: the plane is parallel to one generator and does not pass through the vertex.
Hyperbola: the plane cuts both nappes of the double cone (for example a plane parallel to the axis) and does not pass through the vertex. That is why a hyperbola has two branches.

The four members of the conic family and the cut that produces each one.

One definition for the whole family: eccentricity

Apart from the circle, every conic can also be described with a fixed point F, called the focus, and a fixed line, called the directrix. A point P moves so that the ratio of its distance from the focus to its distance from the directrix stays the same. This constant ratio is the eccentricity $e$ .

e = \frac{∣ P F ∣}{∣ P M ∣} = \frac{distance of P from the focus}{distance of P from the directrix}

M is the foot of the perpendicular from P to the directrix.

Value of $e$	Conic	Shape
$e = 0$	Circle	Perfectly round. It is the limiting case of an ellipse whose two foci meet at the centre.
$0 < e < 1$	Ellipse	Closed oval
$e = 1$	Parabola	One open branch
$e > 1$	Hyperbola	Two open branches

Same focus and directrix, different eccentricity

−4

−3

−2

−1

−4

−3

−2

−1

x

y

directrix x = 2

F

e = 0.5 (ellipse)

e = 1 (parabola)

e = 2 (hyperbola branch)

All three curves have focus

F (0, 0)

and directrix

x = 2

. A larger

e

opens the curve wider: the ellipse is closed, the parabola has one open branch, and for

e > 1

we get a hyperbola.

7.2, 7.2.1Circle: definition and standard formOften examined

A circle is the set of all points in a plane that are at the same distance from a fixed point of the plane. The fixed point is the centre of the circle, and the fixed distance from the centre to any point of the circle is the radius.

Derivation Standard equation of a circle

Aim. Find the equation of the circle with centre $C (h, k)$ and radius $r$ .

−2

−1

−3

−2

−1

x

y

r

C (h, k)

P (x, y)

Every point

P (x, y)

of the circle is at distance

r

from the centre

C (h, k)

Take any point $P (x, y)$ on the circle. By the definition of a circle, its distance from the centre is the radius.
$∣ C P ∣ = r$
Write $∣ C P ∣$ with the distance formula $(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}$ .
$(x - h)^{2} + (y - k)^{2} = r$
Square both sides to remove the square root. Both sides are non-negative, so squaring gives an equivalent equation.
$(x - h)^{2} + (y - k)^{2} = r^{2}$

Result

(x - h)^{2} + (y - k)^{2} = r^{2}

A point lies on the circle exactly when its coordinates satisfy this equation. If the centre is the origin, $h = k = 0$ and the equation becomes $x^{2} + y^{2} = r^{2}$ .

x^{2} + y^{2} = r^{2} (centre at the origin, radius r)

Example 1

Find the equation of the circle with centre at

(2, - 5)

and radius 3 units.

Read the data: the centre is $(h, k) = (2, - 5)$ and $r = 3$ .
Substitute in the standard form $(x - h)^{2} + (y - k)^{2} = r^{2}$ .
$(x - 2)^{2} + (y - (- 5))^{2} = 3^{2}$
Simplify the signs.
$(x - 2)^{2} + (y + 5)^{2} = 9$
If the general form is wanted, expand and collect terms.
$x^{2} - 4 x + 4 + y^{2} + 10 y + 25 - 9 = 0 ⟹ x^{2} + y^{2} - 4 x + 10 y + 20 = 0$

−3

−2

−1

−9

−8

−7

−6

−5

−4

−3

−2

−1

x

y

(2, - 5)

3

(x - 2)^{2} + (y + 5)^{2} = 9

(x - 2)^{2} + (y + 5)^{2} = 9 or x^{2} + y^{2} - 4 x + 10 y + 20 = 0

7.2.2General form of the equation of a circleOften examined

If we expand the standard form, we get an equation containing $x^{2}$ , $y^{2}$ , $x$ , $y$ and a constant. Naming the coefficients $2 g$ , $2 f$ and $c$ gives the general form of the equation of a circle.

Derivation From the standard form to the general form

Aim. Show that $(x - h)^{2} + (y - k)^{2} = r^{2}$ can be written as $x^{2} + y^{2} + 2 g x + 2 f y + c = 0$ .

Expand both squares.
$x^{2} - 2 h x + h^{2} + y^{2} - 2 k y + k^{2} = r^{2}$
Bring every term to the left side.
$x^{2} + y^{2} - 2 h x - 2 k y + (h^{2} + k^{2} - r^{2}) = 0 (i)$
Rename the constants: let $- 2 h = 2 g$ , $- 2 k = 2 f$ and $h^{2} + k^{2} - r^{2} = c$ .
$x^{2} + y^{2} + 2 g x + 2 f y + c = 0$

Result

x^{2} + y^{2} + 2 g x + 2 f y + c = 0

This is the general form. It has equal coefficients of $x^{2}$ and $y^{2}$ and no $x y$ term. It contains three constants $g$ , $f$ , $c$ , so three conditions are needed to fix a circle.

Derivation Theorem 7.1: the general equation represents a circle

Aim. Prove that $x^{2} + y^{2} + 2 g x + 2 f y + c = 0$ represents a circle and find its centre and radius.

Group the $x$ terms and the $y$ terms.
$(x^{2} + 2 g x) + (y^{2} + 2 f y) + c = 0$
Complete each square: add and subtract $g^{2}$ in the $x$ group and $f^{2}$ in the $y$ group, so the equation does not change.
$(x^{2} + 2 g x + g^{2}) + (y^{2} + 2 f y + f^{2}) - g^{2} - f^{2} + c = 0$
Write the perfect squares and move the constants to the right.
$(x + g)^{2} + (y + f)^{2} = g^{2} + f^{2} - c$
Write it in the pattern $(x - h)^{2} + (y - k)^{2} = r^{2}$ .
$[x - (- g)]^{2} + [y - (- f)]^{2} = (g^{2} + f^{2} - c)^{2}$
Compare with the standard form.
$h = - g, k = - f, r = g^{2} + f^{2} - c$

Result

Centre (- g, - f), r = g^{2} + f^{2} - c

Once the coefficients of $x^{2}$ and $y^{2}$ are 1, the centre is minus half the coefficient of $x$ and minus half the coefficient of $y$ . The radius is real only when $g^{2} + f^{2} - c > 0$ .

Value of $g^{2} + f^{2} - c$	What the equation represents
$g^{2} + f^{2} - c > 0$	a real circle with radius $g^{2} + f^{2} - c$
$g^{2} + f^{2} - c = 0$	a point circle: radius 0, only the centre $(- g, - f)$
$g^{2} + f^{2} - c < 0$	an imaginary circle: no real point satisfies the equation

Example 2

Find the centre and radius of the circle

x^{2} + y^{2} - 6 x - 10 y + 18 = 0

Group the terms and complete the squares. Half of $- 6$ is $- 3$ and half of $- 10$ is $- 5$ .
$(x^{2} - 6 x + 9) + (y^{2} - 10 y + 25) - 9 - 25 + 18 = 0$
Write the squares and simplify the constants.
$(x - 3)^{2} + (y - 5)^{2} - 16 = 0 ⟹ (x - 3)^{2} + (y - 5)^{2} = 4^{2}$
Compare with $(x - h)^{2} + (y - k)^{2} = r^{2}$ .
$Centre = (3, 5), r = 4$
Check with the formula: $2 g = - 6$ , $2 f = - 10$ , $c = 18$ , so $g = - 3$ , $f = - 5$ .
$(- g, - f) = (3, 5), r = 9 + 25 - 18 = 16 = 4$

−2

−1

x

y

C (3, 5)

r = 4

x^{2} + y^{2} - 6 x - 10 y + 18 = 0

Centre (3, 5), r = 4

7.2.3Equation of a circle from given conditionsOften examined

A circle has three unknowns: h, k and r in the standard form, or g, f and c in the general form. So we need three independent facts (conditions). Turn each fact into one equation, solve the equations, and substitute back.

Given fact	Equation it gives
The circle passes through (x_1,y_1)	x_1^2+y_1^2+2gx_1+2fy_1+c=0, or \|CP\|=r
The circle passes through A and B	\|CA\|=\|CB\|, so the centre lies on the perpendicular bisector of \overline{AB}
The centre lies on the line ax+by+d=0	ah+bk+d=0
The line L touches the circle at the point P	CP\perp L: (slope of CP)(slope of L)=-1
The line ax+by+d=0 touches the circle	distance from the centre to the line equals r: \dfrac{\|ah+bk+d\|}{\sqrt{a^2+b^2}}=r
The circle touches both axes	\|h\|=\|k\|=r
Two diameters are given	the centre is their point of intersection

Circle through three non-collinear points

Example 3

Find the equation of the circle passing through the points (1,2), (2,3) and (3,5).

Method of the book (using the centre). Let the points be P(1,2), Q(2,3), R(3,5) and let the centre be C(h,k). The centre is equally far from all three points.

|PC|=|QC|. Square both sides and expand; h^2 and k^2 cancel.
(h-1)^2+(k-2)^2=(h-2)^2+(k-3)^2\;\Longrightarrow\;2h+2k=8\;\Longrightarrow\;h+k=4\quad(\text{i})
|PC|=|RC| in the same way.
(h-1)^2+(k-2)^2=(h-3)^2+(k-5)^2\;\Longrightarrow\;4h+6k=29\quad(\text{ii})
Solve: 4\times(i) gives 4h+4k=16. Subtract it from (ii).
2k=13\;\Longrightarrow\;k=\tfrac{13}{2},\qquad h=4-\tfrac{13}{2}=-\tfrac52
Radius r=|CP|.
r^2=\left(1+\tfrac52\right)^2+\left(2-\tfrac{13}{2}\right)^2=\tfrac{49}{4}+\tfrac{81}{4}=\tfrac{130}{4}
Write the standard form and expand.
\left(x+\tfrac52\right)^2+\left(y-\tfrac{13}{2}\right)^2=\tfrac{130}{4}\;\Longrightarrow\;x^2+y^2+5x-13y+\tfrac{25}{4}+\tfrac{169}{4}-\tfrac{130}{4}=0

Quicker method (general form). Put each point in x^2+y^2+2gx+2fy+c=0.

Three equations from the three points.
\begin{aligned}(1,2):&\;\;2g+4f+c=-5\\(2,3):&\;\;4g+6f+c=-13\\(3,5):&\;\;6g+10f+c=-34\end{aligned}
Subtract the first from the second and the second from the third to remove c.
2g+2f=-8,\qquad 2g+4f=-21
Solve these two, then find c.
f=-\tfrac{13}{2},\quad g=\tfrac52,\quad c=-5-5+26=16

x^2+y^2+5x-13y+16=0,\qquad\text{centre }\left(-\tfrac52,\tfrac{13}{2}\right),\ r=\tfrac{\sqrt{130}}{2}

Two points and the centre on a given line

Example 4

Find the equation of the circle passing through the points (1,0) and (0,1) and having its centre on the line x-2y+3=0.

Let the centre be C(h,k). It lies on the given line.
h-2k+3=0\quad(\text{i})
The centre is equally far from P(1,0) and Q(0,1).
(h-1)^2+k^2=h^2+(k-1)^2\;\Longrightarrow\;-2h+2k=0\;\Longrightarrow\;h=k\quad(\text{ii})
Put h=k in (i).
k-2k+3=0\;\Longrightarrow\;k=3,\quad h=3
Radius r=|CP|.
r=\sqrt{(3-1)^2+(3-0)^2}=\sqrt{13}
Equation of the circle.
(x-3)^2+(y-3)^2=13\;\Longrightarrow\;x^2+y^2-6x-6y+5=0

x^2+y^2-6x-6y+5=0

Two points and the tangent at one of them

Example 5

Find the equation of the circle passing through the point (2,3) if the line x+y-4=0 is the tangent to the circle at (3,1).

Let the centre be C(h,k), with Q(2,3) and the point of contact P(3,1). Both are on the circle, so |CP|=|CQ|.
(h-3)^2+(k-1)^2=(h-2)^2+(k-3)^2\;\Longrightarrow\;2h-4k=-3\quad(\text{i})
The radius CP is perpendicular to the tangent x+y-4=0, whose slope is -1.
\left(\dfrac{k-1}{h-3}\right)(-1)=-1\;\Longrightarrow\;k-1=h-3\;\Longrightarrow\;h-k=2\quad(\text{ii})
Solve (i) and (ii): from (ii) h=k+2, so 2k+4-4k=-3.
k=\tfrac72,\qquad h=\tfrac{11}{2}
Radius r=|CP|.
r^2=\left(\tfrac{11}{2}-3\right)^2+\left(\tfrac72-1\right)^2=\tfrac{25}{4}+\tfrac{25}{4}=\tfrac{50}{4},\qquad r=\tfrac{5\sqrt2}{2}
Equation of the circle.
\left(x-\tfrac{11}{2}\right)^2+\left(y-\tfrac72\right)^2=\tfrac{50}{4}\;\Longrightarrow\;x^2+y^2-11x-7y+30=0

x^2+y^2-11x-7y+30=0

Two points and a line touching the circle

Example 6

Find the equation of the circle passing through the points (-1,2), (3,2) and touching the line x-2y+1=0.

Let the centre be C(h,k). Equal distances from P(-1,2) and Q(3,2):
(h+1)^2+(k-2)^2=(h-3)^2+(k-2)^2\;\Longrightarrow\;8h=8\;\Longrightarrow\;h=1\quad(\text{i})
The line touches the circle, so the distance from C to the line equals the radius |CP|.
\sqrt{(h+1)^2+(k-2)^2}=\dfrac{|h-2k+1|}{\sqrt{1^2+(-2)^2}}\quad(\text{ii})
Put h=1 in (ii) and square both sides.
4+(k-2)^2=\dfrac{(2-2k)^2}{5}
Multiply by 5 and simplify.
5k^2-20k+40=4k^2-8k+4\;\Longrightarrow\;k^2-12k+36=0\;\Longrightarrow\;(k-6)^2=0
So k=6 (a double root). The radius is |CP|.
r=\sqrt{(1+1)^2+(6-2)^2}=\sqrt{20}
Equation of the circle.
(x-1)^2+(y-6)^2=20\;\Longrightarrow\;x^2+y^2-2x-12y+17=0

The circle touches the line x-2y+1=0 exactly at Q(3,2), because Q lies on that line. The radius CQ is perpendicular to the line.

x^2+y^2-2x-12y+17=0

Other useful forms

Circle through the intersection of two circles S_1=0 and S_2=0: S_1+\lambda S_2=0 with \lambda\neq-1. Find \lambda from one more condition (used in Exercise 7.1 Q11).
Circle on a diameter with ends A(x_1,y_1) and B(x_2,y_2): (x-x_1)(x-x_2)+(y-y_1)(y-y_2)=0. Reason: for any other point P of the circle, \angle APB=90^\circ, so (slope of PA)(slope of PB)=-1.
Circle touching both axes with radius r in the first quadrant: centre (r,r), so (x-r)^2+(y-r)^2=r^2. A circle touching only the x-axis has |k|=r; touching only the y-axis has |h|=r.
Circle through the origin: c=0, so x^2+y^2+2gx+2fy=0.

7.3Line and a circleOften examined

A line and a circle can meet in two points, touch at exactly one point, or not meet at all. The fastest test compares the perpendicular distance d from the centre to the line with the radius r. A second test substitutes the line into the circle and looks at the discriminant \Delta=B^2-4AC of the quadratic that results.

Position of the line	Distance test	Discriminant test
Cuts the circle at two distinct points (a secant)	d<r	\Delta>0
Touches the circle at one point (a tangent)	d=r	\Delta=0
Neither touches nor intersects the circle	d>r	\Delta<0

Parallel lines with slope \tfrac12. The distance from the centre decides: d<r gives two points, d=r one point (a tangent), d>r none.

Derivation Conditions for the line y=mx+c and the circle x^2+y^2=r^2

Aim. Find when the line y=mx+c cuts, touches or misses the circle x^2+y^2=r^2.

The centre of the circle is (0,0) and its radius is r. Write the line in general form.
mx-y+c=0
Distance of the centre (0,0) from the line.
d=\dfrac{|m(0)-0+c|}{\sqrt{m^2+(-1)^2}}=\dfrac{|c|}{\sqrt{m^2+1}}
Two points: d<r. Multiply by \sqrt{m^2+1} and square (both sides are positive).
|c|<r\sqrt{m^2+1}\;\Longrightarrow\;c^2<r^2\left(m^2+1\right)
Tangent: d=r.
|c|=r\sqrt{m^2+1}\;\Longrightarrow\;c=\pm r\sqrt{m^2+1}
No common point: d>r.
c^2>r^2\left(m^2+1\right)
Put the tangent value of c into y=mx+c.
y=mx\pm r\sqrt{m^2+1}

Result\begin{aligned}c^2&=r^2\left(1+m^2\right)&&\text{(condition of tangency)}\\ y&=mx\pm r\sqrt{1+m^2}&&\text{(tangents in slope form)}\end{aligned}

For every slope m there are exactly two tangents to a circle, one on each side of the centre. They are parallel to each other.

The same result from the discriminant

Substitute y=mx+c into x^2+y^2=r^2.
\left(1+m^2\right)x^2+2mcx+\left(c^2-r^2\right)=0
Its discriminant is
\Delta=4m^2c^2-4\left(1+m^2\right)\left(c^2-r^2\right)=4\left[r^2\left(1+m^2\right)-c^2\right]
So \Delta>0, \Delta=0, \Delta<0 give exactly the same three conditions. When \Delta=0 the equal roots give the point of contact.
\left(\dfrac{-mc}{1+m^2},\ \dfrac{c}{1+m^2}\right)=\left(\mp\dfrac{rm}{\sqrt{1+m^2}},\ \pm\dfrac{r}{\sqrt{1+m^2}}\right)

Example 7

Check whether the line x+2y-3=0 is a tangent to the circle x^2+y^2=16 or not.

The circle x^2+y^2=16=4^2 has centre (0,0) and r=4. Write the line in slope form.
2y=-x+3\;\Longrightarrow\;y=-\tfrac12x+\tfrac32,\qquad m=-\tfrac12,\ c=\tfrac32
For a tangent we need |c|=r\sqrt{m^2+1}. Work out the right side.
r\sqrt{m^2+1}=4\sqrt{\tfrac14+1}=4\cdot\dfrac{\sqrt5}{2}=2\sqrt5\approx4.47
Compare: |c|=1.5\neq2\sqrt5, so the line is not a tangent. In fact c^2=\tfrac94<r^2(m^2+1)=20, so the line cuts the circle at two points.
Distance check: the distance of the centre from the line is less than the radius.
d=\dfrac{|0+0-3|}{\sqrt{1+4}}=\dfrac{3}{\sqrt5}\approx1.34<4

\text{Not a tangent: the line cuts the circle at two points.}

Example (tangents with a given slope)

Find the tangents to the circle x^2+y^2=9 that have slope 2.

Here r=3 and m=2. Use y=mx\pm r\sqrt{1+m^2}.
y=2x\pm3\sqrt{1+4}=2x\pm3\sqrt5
So the value of c for which y=2x+c touches the circle is c=\pm3\sqrt5 (this was a 2026 MCQ).

y=2x\pm3\sqrt5

7.4Tangent and normal to a circle at a pointOften examined

The tangent to a circle at a point P of the circle is the line that touches the circle at P only. The normal at P is the line through P perpendicular to the tangent. For a circle, the normal always passes through the centre, because the tangent is perpendicular to the radius.

Derivation Equation of the tangent at P(x_1,y_1)

Aim. Find the tangent at the point P(x_1,y_1) of the circle x^2+y^2+2gx+2fy+c=0.

The slope of the tangent is the derivative \dfrac{dy}{dx} at P. Differentiate the equation of the circle with respect to x (implicit differentiation).
2x+2y\dfrac{dy}{dx}+2g+2f\dfrac{dy}{dx}=0
Make \dfrac{dy}{dx} the subject.
(2y+2f)\dfrac{dy}{dx}=-(2x+2g)\;\Longrightarrow\;\dfrac{dy}{dx}=-\dfrac{x+g}{y+f}
At P(x_1,y_1) this is the slope m of the tangent.
m=-\dfrac{x_1+g}{y_1+f}
Point-slope form y-y_1=m(x-x_1); multiply by (y_1+f).
(y_1+f)(y-y_1)=-(x_1+g)(x-x_1)
Expand and collect all terms on one side.
(x_1+g)x+(y_1+f)y-\left(x_1^2+y_1^2+gx_1+fy_1\right)=0\qquad(\text{i})
P lies on the circle, so x_1^2+y_1^2+2gx_1+2fy_1+c=0. Rearranged:
x_1^2+y_1^2+gx_1+fy_1=-\left(gx_1+fy_1+c\right)
Substitute this in (i).
(x_1+g)x+(y_1+f)y+\left(gx_1+fy_1+c\right)=0

Result\begin{aligned}&(x_1+g)x+(y_1+f)y+\left(gx_1+fy_1+c\right)=0\\ \Longleftrightarrow\ &xx_1+yy_1+g(x+x_1)+f(y+y_1)+c=0\end{aligned}

For the circle x^2+y^2=r^2 (g=f=0, c=-r^2) the tangent at (x_1,y_1) is simply xx_1+yy_1=r^2.

Derivation Equation of the normal at P(x_1,y_1)

Aim. Find the normal at P(x_1,y_1) of the circle x^2+y^2+2gx+2fy+c=0.

The normal is perpendicular to the tangent, so its slope is the negative reciprocal of m=-\dfrac{x_1+g}{y_1+f}.
m_1=-\dfrac1m=\dfrac{y_1+f}{x_1+g}
Point-slope form through P(x_1,y_1); multiply by (x_1+g).
(x_1+g)(y-y_1)=(y_1+f)(x-x_1)
Expand; the terms x_1y_1 cancel.
(y_1+f)x-(x_1+g)y-\left(fx_1-gy_1\right)=0
Check: put the centre (-g,-f) in the left side.
(y_1+f)(-g)-(x_1+g)(-f)-fx_1+gy_1=0

Result(y_1+f)x-(x_1+g)y-\left(fx_1-gy_1\right)=0

The normal at any point of a circle passes through the centre. So the quickest way to write the normal is: the line through P and the centre.

The tangent is perpendicular to the radius

The radius through P(x_1,y_1) joins the centre C(-g,-f) to P, so its slope is \dfrac{y_1+f}{x_1+g}. The slope of the tangent is -\dfrac{x_1+g}{y_1+f}. Their product is -1, so the tangent is perpendicular to the radius CP. If y_1+f=0, the radius is horizontal and the tangent is the vertical line x=x_1; if x_1+g=0, the radius is vertical and the tangent is horizontal.

Example 8

Find the equations of the tangent and normal to the circle x^2+y^2-4x+2y-5=0 at the point P(1,2).

Method 1 (differentiation).

Differentiate with respect to x.
2x+2y\dfrac{dy}{dx}-4+2\dfrac{dy}{dx}=0\;\Longrightarrow\;\dfrac{dy}{dx}=\dfrac{-2x+4}{2y+2}=\dfrac{-x+2}{y+1}
Slope of the tangent at P(1,2).
m=\dfrac{-1+2}{2+1}=\dfrac13
Tangent by the point-slope form.
y-2=\tfrac13(x-1)\;\Longrightarrow\;3y-6=x-1\;\Longrightarrow\;x-3y+5=0
The normal has slope -3.
y-2=-3(x-1)\;\Longrightarrow\;3x+y-5=0

Method 2 (formulas). Compare with x^2+y^2+2gx+2fy+c=0: g=-2, f=1, c=-5, and x_1=1, y_1=2.

Tangent: (x_1+g)x+(y_1+f)y+(gx_1+fy_1+c)=0.
(1-2)x+(2+1)y+(-2+2-5)=0\;\Longrightarrow\;-x+3y-5=0\;\Longrightarrow\;x-3y+5=0
Normal: (y_1+f)x-(x_1+g)y-(fx_1-gy_1)=0.
(2+1)x-(1-2)y-(1+4)=0\;\Longrightarrow\;3x+y-5=0

The normal at P passes through the centre C; the tangent is perpendicular to it.

\text{Tangent: }x-3y+5=0,\qquad\text{Normal: }3x+y-5=0

Two circles that touch

Let two circles have centres C_1, C_2 and radii r_1, r_2. They touch externally when |C_1C_2|=r_1+r_2 and touch internally when |C_1C_2|=|r_1-r_2|. At the point of contact they have a common tangent. If both equations are written with x^2 and y^2 having coefficient 1, subtracting them removes x^2 and y^2 and leaves the common tangent at the point of contact:

S_1-S_2=0\qquad\text{(common tangent at the point of contact of two touching circles)}

For external contact, the point of contact divides C_1C_2 internally in the ratio r_1:r_2 (see Exercise 7.2 Q7).

7.5Position of a point with respect to a circle

A point P is outside, on or inside a circle when its distance from the centre is greater than, equal to or less than the radius. This comparison becomes a very short test: substitute the point into the left side of the equation of the circle.

Derivation Test for the position of P(x_1,y_1)

Aim. Decide whether P(x_1,y_1) lies outside, on or inside the circle x^2+y^2+2gx+2fy+c=0.

The centre is C(-g,-f) and the radius is r=\sqrt{g^2+f^2-c}. Compare |CP| with r.
|CP|\ \gtreqless\ r
Write |CP| with the distance formula.
\sqrt{(x_1+g)^2+(y_1+f)^2}\ \gtreqless\ \sqrt{g^2+f^2-c}
Square both sides (both are positive) and expand.
x_1^2+2gx_1+g^2+y_1^2+2fy_1+f^2\ \gtreqless\ g^2+f^2-c
Cancel g^2 and f^2 and move everything to the left.
x_1^2+y_1^2+2gx_1+2fy_1+c\ \gtreqless\ 0

ResultS_1=x_1^2+y_1^2+2gx_1+2fy_1+c\;\begin{cases}>0 & \text{P outside}\\=0 & \text{P on the circle}\\<0 & \text{P inside}\end{cases}

S_1 is the value of the left side of the circle's equation at P (with the coefficients of x^2 and y^2 equal to 1). Its sign alone tells us where P is. In fact S_1=|CP|^2-r^2.

Example 9

Check whether the point P(2,4) lies outside, on or inside the circle 2x^2+2y^2-6x+8y+1=0.

Make the coefficients of x^2 and y^2 equal to 1 by dividing by 2.
x^2+y^2-3x+4y+\tfrac12=0
Substitute x_1=2, y_1=4 in the left side.
S_1=2^2+4^2-3(2)+4(4)+\tfrac12=4+16-6+16+\tfrac12=\tfrac{61}{2}
S_1>0, so the point lies outside the circle.

S_1=\tfrac{61}{2}>0:\ P\text{ lies outside the circle}

7.6Length of a tangent from a point outside the circleOften examined

From a point P outside a circle we can draw exactly two tangents, touching the circle at A and B. The length of the tangent is the distance |PA| from P to the point of contact. Both tangents have the same length.

Derivation Length of the tangent

Aim. Find |PA| when P(x_1,y_1) lies outside the circle x^2+y^2+2gx+2fy+c=0.

The tangents PA and PB from an external point are equal. Each radius meets its tangent at a right angle.

The centre is C(-g,-f) and r=\sqrt{g^2+f^2-c}. The radius CA is perpendicular to the tangent PA, so triangle PAC has a right angle at A. By Pythagoras:
|CA|^2+|AP|^2=|CP|^2
Put |CA|=r and write |CP| with the distance formula.
\left(g^2+f^2-c\right)+|AP|^2=(x_1+g)^2+(y_1+f)^2
Expand the right side.
g^2+f^2-c+|AP|^2=x_1^2+2gx_1+g^2+y_1^2+2fy_1+f^2
Cancel g^2 and f^2 and make |AP|^2 the subject.
|AP|^2=x_1^2+y_1^2+2gx_1+2fy_1+c

Result|AP|=\sqrt{x_1^2+y_1^2+2gx_1+2fy_1+c}=\sqrt{S_1}

The square of the tangent length is the value of the circle's left side at P. For x^2+y^2=r^2 it is \sqrt{x_1^2+y_1^2-r^2}. The formula needs the coefficients of x^2 and y^2 to be 1.

Example 10

Find the length of the tangent line to the circle x^2+y^2+3x-4y+15=0 from the point (1,2).

The coefficients of x^2 and y^2 are already 1. Substitute (1,2) into \sqrt{x_1^2+y_1^2+3x_1-4y_1+15}.
|PT|=\sqrt{1+4+3-8+15}=\sqrt{15}\text{ units}

\sqrt{15}\text{ units (formula only)}

Example (2026 MCQ)

Find the length of the tangent from the point (10,0) to the circle x^2+y^2=36.

Here S_1=x_1^2+y_1^2-36.
|PT|=\sqrt{10^2+0^2-36}=\sqrt{64}=8
Check with Pythagoras: the distance from the centre is 10 and the radius is 6, so \sqrt{10^2-6^2}=8.

8\text{ units}

Tangents from an external point and useful results

**Equations of the tangents from P(x_1,y_1):** write y-y_1=m(x-x_1), set the distance from the centre equal to r, and solve the quadratic for m. If only one value of m appears, the other tangent is the vertical line x=x_1.
Points of contact: each is the foot of the perpendicular from the centre to a tangent, or solve the tangent with the circle (a double root).
Equal tangents: |PA|=|PB|=\sqrt{S_1}, and CP bisects the angle APB.
Tangent and secant: if a line through P cuts the circle at B and D, then |PA|^2=|PB|\cdot|PD| (Exercise 7.3 Q7).
Circumscribed quadrilateral: if the four sides of ABCD touch a circle, then AB+CD=AD+BC, because the two tangents from each vertex are equal (Exercise 7.3 Q5).

7.7, 7.7.1Parabola: definition and standard equationOften examined

A parabola is the set of all points in a plane that are equally distant from a fixed point and a fixed line of the plane (the fixed point is not on the fixed line). The fixed point is the focus and the fixed line is the directrix.

The eccentricity is the ratio (distance from the focus) : (distance from the directrix). For every point of a parabola these two distances are equal, so the eccentricity of a parabola is e=1.

Derivation Standard equation y^2=4ax

Aim. Find the equation of the parabola with focus F(a,0), a>0, and directrix x=-a (that is, x+a=0).

For every point P of the parabola, |PF|=|PM|: its distance from the focus equals its distance from the directrix.

Take any point P(x,y) on the parabola and let M be the foot of the perpendicular from P to the directrix. By the definition,
|PF|=|PM|
Use the distance formula for |PF| and the point-to-line distance for |PM|.
\sqrt{(x-a)^2+(y-0)^2}=\dfrac{|x+a|}{\sqrt{1^2+0^2}}=|x+a|
Square both sides.
x^2-2ax+a^2+y^2=x^2+2ax+a^2
Cancel x^2 and a^2 and collect terms.
y^2=4ax

Resulty^2=4ax

The curve is symmetric about the x-axis (replacing y by -y gives the same equation), passes through the origin, and exists only for x\ge0, so it opens to the right.

Placing the focus on the other half-axes in the same way gives the other three standard forms.

In every case a>0, the vertex is (0,0) and the length of the latus rectum is 4a.
Equation	Focus	Directrix	Axis	Opens	Ends of latus rectum
y^2=4ax	(a,0)	x=-a	y=0	right	(a,\pm2a)
y^2=-4ax	(-a,0)	x=a	y=0	left	(-a,\pm2a)
x^2=4ay	(0,a)	y=-a	x=0	upward	(\pm2a,a)
x^2=-4ay	(0,-a)	y=a	x=0	downward	(\pm2a,-a)

The four standard parabolas drawn with a=1. Each focus lies inside its own curve, a distance a from the vertex O.

7.7.2Elements of a parabolaOften examined

The following names are used for every parabola. The examples refer to y^2=4ax.

Directrix: The fixed line of the definition: x=-a.
Focus: The fixed point of the definition, not lying on the directrix: F(a,0).
Axis (axis of symmetry): The line through the focus perpendicular to the directrix: y=0. The parabola is symmetric about it.
Vertex: The point where the parabola cuts its axis: O(0,0). It is the point of the parabola closest to the focus, midway between the focus and the directrix.
Chord: A line segment whose end points are on the parabola.
Focal chord: A chord that passes through the focus.
Focal distance: The distance of a point of the parabola from the focus.
Latus rectum: The focal chord parallel to the directrix (perpendicular to the axis). Its length is 4a and its ends are (a,\pm2a).

Focal distance of a point

If P(x_1,y_1) is on y^2=4ax, its distance from the focus equals its distance from the directrix x=-a, which is x_1+a. So the focal distance is found without a square root:

|PF|=x_1+a\qquad\text{(for }y^2=4ax)

Parametric form and focal chords

The point (at^2,\,2at) lies on y^2=4ax for every real t, because (2at)^2=4a(at^2). The number t is called the parameter of the point.

Derivation Condition for a focal chord

Aim. Show that the chord joining P(at_1^2,2at_1) and Q(at_2^2,2at_2) passes through the focus when t_1t_2=-1.

Slope of PQ.
m=\dfrac{2a(t_2-t_1)}{a(t_2^2-t_1^2)}=\dfrac{2}{t_1+t_2}
Equation of PQ through P.
y-2at_1=\dfrac{2}{t_1+t_2}\left(x-at_1^2\right)
Put the focus (a,0) in this equation and simplify.
-2at_1(t_1+t_2)=2a-2at_1^2\;\Longrightarrow\;-2at_1t_2=2a
Divide by -2a.
t_1t_2=-1

ResultPQ\text{ is a focal chord}\iff t_1t_2=-1

The ends of a focal chord have parameters whose product is -1. For example, the ends of the latus rectum have t=1 and t=-1.

7.7.3Parabola with vertex at (h, k)Often examined

Moving the vertex from the origin to V(h,k) without turning the parabola replaces x by x-h and y by y-k. Every element moves by the same amount.

Vertex V(h,k), a>0, length of latus rectum 4a in every case.
Equation	Opens	Focus	Directrix	Axis	Ends of latus rectum
(y-k)^2=4a(x-h)	right	(h+a,\,k)	x=h-a	y=k	(h+a,\,k\pm2a)
(y-k)^2=-4a(x-h)	left	(h-a,\,k)	x=h+a	y=k	(h-a,\,k\pm2a)
(x-h)^2=4a(y-k)	upward	(h,\,k+a)	y=k-a	x=h	(h\pm2a,\,k+a)
(x-h)^2=-4a(y-k)	downward	(h,\,k-a)	y=k+a	x=h	(h\pm2a,\,k-a)

Vertex (1,2), 4a=8 so a=2, focus (3,2), directrix x=-1, axis y=2, latus rectum from (3,-2) to (3,6).

7.7.4Length of the latus rectum

Derivation Length of the latus rectum

Aim. Show that the latus rectum of y^2=4ax has length 4a.

The latus rectum \overline{AB} passes through the focus F(a,0) and is perpendicular to the axis. Let its length be l. F is its midpoint, so |FA|=\tfrac{l}{2} and
A=\left(a,\ \tfrac{l}{2}\right)
A lies on the parabola, so its coordinates satisfy y^2=4ax.
\left(\tfrac{l}{2}\right)^2=4a(a)\;\Longrightarrow\;\dfrac{l^2}{4}=4a^2\;\Longrightarrow\;l^2=16a^2
A length is positive, so take the positive root.
l=4a

Result\text{Length of latus rectum}=4a

The latus rectum is 4 times the distance from the vertex to the focus. Its ends are (a,2a) and (a,-2a), which helps to sketch the width of the parabola at the focus.

Example 11

Find the equation of the parabola with focus at (2,5) and directrix x+8=0.

Let P(x,y) be any point of the parabola. By the definition |PF|=|PM|, where |PM| is the distance from P to the directrix.
\sqrt{(x-2)^2+(y-5)^2}=\dfrac{|x+8|}{\sqrt{1^2+0^2}}
Square both sides and keep the y terms on the left.
(y-5)^2=(x+8)^2-(x-2)^2
Expand the right side: x^2+16x+64-x^2+4x-4.
(y-5)^2=20x+60
Factor to see the elements.
(y-5)^2=20(x+3)=4(5)(x+3)

So the vertex is (-3,5) (midway between the focus and the directrix), a=5, the parabola opens to the right, and the latus rectum has length 20.

(y-5)^2=20(x+3)\quad\text{or}\quad y^2-10y-20x-35=0

7.7.5Elements of a parabola from its equation, and its graphOften examined

To find the elements of a parabola from its equation, first bring the equation to one of the standard forms, then compare.

Keep the squared variable and its first-degree term on one side and complete the square.
Write the result as (x-h)^2=\pm4a(y-k) or (y-k)^2=\pm4a(x-h) and read h, k and a.
Put X=x-h, Y=y-k; write the vertex, focus, axis, directrix and latus rectum in X,Y; then return to x,y.
For the graph, also find the intercepts and plot the ends of the latus rectum.

Example 12

Find the elements of the parabola y=x^2-3x+7 and draw its graph.

Complete the square in x: add \left(\tfrac32\right)^2=\tfrac94 to both sides of x^2-3x=y-7.
x^2-3x+\tfrac94=y-7+\tfrac94\;\Longrightarrow\;\left(x-\tfrac32\right)^2=y-\tfrac{19}{4}
Let X=x-\tfrac32 and Y=y-\tfrac{19}{4}. Then X^2=Y=4\left(\tfrac14\right)Y, the form X^2=4aY with
a=\tfrac14
Vertex: X=0, Y=0.
x=\tfrac32,\ y=\tfrac{19}{4}\;\Longrightarrow\;V\left(\tfrac32,\tfrac{19}{4}\right)
Focus: X=0, Y=a=\tfrac14.
x=\tfrac32,\ y=\tfrac{19}{4}+\tfrac14=5\;\Longrightarrow\;F\left(\tfrac32,5\right)
Axis: X=0.
x=\tfrac32
Directrix: Y=-a.
y-\tfrac{19}{4}=-\tfrac14\;\Longrightarrow\;y=\tfrac92
Latus rectum: length 4a=1; its ends are X=\pm2a=\pm\tfrac12, Y=a.
(1,5)\ \text{and}\ (2,5)
Intercepts: y=0 gives x^2-3x+7=0 with discriminant 9-28=-19<0, so there is no x-intercept. x=0 gives y=7.
y\text{-intercept }(0,7)

Vertex V\left(\tfrac32,\tfrac{19}{4}\right), focus F\left(\tfrac32,5\right), directrix y=\tfrac92, axis x=\tfrac32, latus rectum from (1,5) to (2,5), y-intercept (0,7), no x-intercept.

V\left(\tfrac32,\tfrac{19}{4}\right),\ F\left(\tfrac32,5\right),\ x=\tfrac32,\ y=\tfrac92,\ \text{LR}=1

7.7.6General equation of a parabolaOften examined

Derivation y=ax^2+bx+c is a parabola

Aim. Prove that y=ax^2+bx+c, where a,b,c are real and a\neq0, represents a parabola, and find its vertex.

Keep the x terms on one side.
ax^2+bx=y-c
Divide by a (allowed because a\neq0).
x^2+\dfrac{b}{a}x=\dfrac1a(y-c)
Complete the square: add \dfrac{b^2}{4a^2} to both sides.
\left(x+\dfrac{b}{2a}\right)^2=\dfrac1a(y-c)+\dfrac{b^2}{4a^2}=\dfrac1a\left[y+\dfrac{b^2-4ac}{4a}\right]
Write \dfrac1a as 4\left(\dfrac{1}{4a}\right) and arrange in the pattern (x-h)^2=4p(y-k).
\left[x-\left(-\dfrac{b}{2a}\right)\right]^2=4\left(\dfrac{1}{4a}\right)\left[y-\dfrac{4ac-b^2}{4a}\right]

Result(x-h)^2=4p(y-k),\quad h=-\dfrac{b}{2a},\quad k=\dfrac{4ac-b^2}{4a},\quad p=\dfrac{1}{4a}

The graph is a parabola with vertex \left(-\dfrac{b}{2a},\ \dfrac{4ac-b^2}{4a}\right) and vertical axis. It opens upward when a>0 (p>0) and downward when a<0 (p<0). In the same way x=ay^2+by+c is a parabola that opens right or left.

Example 13

Find the equation of the parabola with vertex at (2,3) and focus at (7,3).

The vertex V(2,3) and focus F(7,3) have the same y-coordinate, so the axis is the horizontal line y=3. The focus is to the right of the vertex, so the parabola opens to the right: (y-k)^2=4a(x-h).
a is the distance from the vertex to the focus.
a=|VF|=\sqrt{(7-2)^2+(3-3)^2}=5
Substitute h=2, k=3, a=5.
(y-3)^2=4(5)(x-2)=20(x-2)
Other elements: directrix x=h-a=-3, latus rectum 4a=20.

(y-3)^2=20(x-2)

Example 14

Find the equation of the parabola with focus at (3,-1) and directrix 3x-4y+1=0.

Let P(x,y) be on the parabola. By the definition |PF|=|PM|, with |PM| the distance from P to the directrix.
\sqrt{(x-3)^2+(y+1)^2}=\dfrac{|3x-4y+1|}{\sqrt{3^2+(-4)^2}}=\dfrac{|3x-4y+1|}{5}
Square both sides and multiply by 25.
25\left[x^2-6x+9+y^2+2y+1\right]=9x^2+16y^2+1-24xy+6x-8y
Expand the left side and bring every term to the left.
25x^2-150x+25y^2+50y+250-9x^2-16y^2-1+24xy-6x+8y=0
Collect like terms.
16x^2+24xy+9y^2-156x+58y+249=0

The directrix is slanted, so the axis is slanted too and the equation contains an xy term.

16x^2+24xy+9y^2-156x+58y+249=0

Example 15

Find the equation of the parabola with vertex at (2,3) and directrix y=-4.

The directrix y=-4 is horizontal and below the vertex. The focus is on the opposite side of the vertex, so the parabola opens upward: (x-h)^2=4a(y-k).
a is the distance from the vertex (2,3) to the line y+4=0.
a=\dfrac{|3+4|}{\sqrt{0^2+1^2}}=7
Substitute h=2, k=3, a=7.
(x-2)^2=4(7)(y-3)=28(y-3)
The focus is (h,k+a)=(2,10).

(x-2)^2=28(y-3)

7.8, 7.8.1Condition for a line to be tangent to a parabolaOften examined

Substituting a line into the equation of a parabola gives a quadratic equation. Two different roots mean two points of intersection. The line is a tangent when the two roots are equal, that is when the discriminant is zero.

Derivation Condition for y=mx+c to touch (y-k)^2=4a(x-h)

Aim. Find the value of c for which the line y=mx+c is a tangent to the parabola (y-k)^2=4a(x-h).

Put y=mx+c in the equation of the parabola.
\big[mx+(c-k)\big]^2=4a(x-h)
Expand and arrange as a quadratic in x.
m^2x^2+\big[2m(c-k)-4a\big]x+(c-k)^2+4ah=0
For a tangent the two roots must be equal, so the discriminant is zero.
\big[2m(c-k)-4a\big]^2-4m^2\left[(c-k)^2+4ah\right]=0
Expand. The terms 4m^2(c-k)^2 cancel.
-16am(c-k)+16a^2-16ahm^2=0\;\Longrightarrow\;-16a\left[m(c-k)-a+hm^2\right]=0
Since a\neq0, the bracket is zero. Solve for c.
m(c-k)=a-m^2h\;\Longrightarrow\;c=\dfrac{a-m^2h+mk}{m}
Put this c in y=mx+c.
y=mx+\dfrac{a-m^2h+mk}{m}=mx-\dfrac{m^2h-mk-a}{m}

Resultc=\dfrac{a-m^2h+mk}{m},\qquad\text{tangent: }y-k=m(x-h)+\dfrac{a}{m}

For the vertex at the origin (h=k=0) the condition is c=\dfrac{a}{m}, so y=mx+\dfrac{a}{m} is a tangent to y^2=4ax for every slope m\neq0. It touches the parabola at \left(\dfrac{a}{m^2},\dfrac{2a}{m}\right).

Each row comes from the same discriminant argument.
Parabola	Condition for y=mx+c to be a tangent	Tangent in slope form
y^2=4ax	c=\dfrac{a}{m}	y=mx+\dfrac{a}{m}
y^2=-4ax	c=-\dfrac{a}{m}	y=mx-\dfrac{a}{m}
x^2=4ay	c=-am^2	y=mx-am^2
x^2=-4ay	c=am^2	y=mx+am^2

Example 16

Find the equation of the tangent to the parabola y^2-6y-16x+25=0 with slope \tfrac12.

Complete the square in y.
y^2-6y+9=16x-25+9\;\Longrightarrow\;(y-3)^2=16x-16=16(x-1)
Compare with (y-k)^2=4a(x-h).
h=1,\quad k=3,\quad 4a=16\Rightarrow a=4,\qquad m=\tfrac12
Use y-k=m(x-h)+\dfrac{a}{m}.
y-3=\tfrac12(x-1)+\dfrac{4}{1/2}=\tfrac12x-\tfrac12+8
Simplify and multiply by 2.
y=\tfrac12x+\tfrac{21}{2}\;\Longrightarrow\;x-2y+21=0
Check: put x=2y-21 in the parabola. The quadratic is a perfect square, so the line touches it at one point.
y^2-6y-16(2y-21)+25=y^2-38y+361=(y-19)^2

The corrected line x-2y+21=0 touches the parabola at (17,19). The book's line x-2y+18=0 cuts it at two points.

x-2y+21=0

7.8.2, 7.8.3Tangent and normal to a parabola at a pointOften examined

Derivation Tangent at P(x_1,y_1)

Aim. Find the tangent at the point P(x_1,y_1) of the parabola (y-k)^2=4a(x-h).

Differentiate the equation with respect to x.
2(y-k)\dfrac{dy}{dx}=4a\;\Longrightarrow\;\dfrac{dy}{dx}=\dfrac{2a}{y-k}
At P(x_1,y_1) the slope of the tangent is
m=\dfrac{2a}{y_1-k}
Point-slope form, multiplied by (y_1-k).
(y_1-k)(y-y_1)=2a(x-x_1)
Expand; this is the book's form.
2ax-(y_1-k)y-2ax_1+(y_1-k)y_1=0
Vertex at the origin (h=k=0): use y_1^2=4ax_1 to simplify.
2ax-y_1y-2ax_1+4ax_1=0\;\Longrightarrow\;yy_1=2a(x+x_1)

Result(y_1-k)(y-y_1)=2a(x-x_1);\qquad yy_1=2a(x+x_1)\ \text{ for }y^2=4ax

The replacement rule works for parabolas too: in y^2=4ax replace y^2 by yy_1 and x by \tfrac12(x+x_1).

Derivation Normal at P(x_1,y_1)

Aim. Find the normal at P(x_1,y_1) of the parabola (y-k)^2=4a(x-h).

The normal is perpendicular to the tangent, so its slope is
m_1=-\dfrac{1}{m}=-\dfrac{y_1-k}{2a}
Point-slope form through P.
y-y_1=-\dfrac{y_1-k}{2a}(x-x_1)
Multiply by 2a and collect terms.
(y_1-k)(x-x_1)+2a(y-y_1)=0

Result\begin{aligned}(y_1-k)(x-x_1)+2a(y-y_1)&=0\\ y_1(x-x_1)+2a(y-y_1)&=0\ \text{ for }{y^2=4ax}\end{aligned}

The normal at the vertex is the axis itself, and the tangent at the vertex is perpendicular to the axis.

Parabola	Tangent at (x_1,y_1)	Normal at (x_1,y_1)
y^2=4ax	yy_1=2a(x+x_1)	y_1(x-x_1)+2a(y-y_1)=0
y^2=-4ax	yy_1=-2a(x+x_1)	y_1(x-x_1)-2a(y-y_1)=0
x^2=4ay	xx_1=2a(y+y_1)	x_1(y-y_1)+2a(x-x_1)=0
x^2=-4ay	xx_1=-2a(y+y_1)	x_1(y-y_1)-2a(x-x_1)=0

Example 17

Find the equations of the tangent and normal to the parabola y^2-6y+8x-9=0 at the point P\left(\tfrac14,-1\right).

Check that P is on the curve: 1+6+2-9=0. Differentiate with respect to x.
2y\dfrac{dy}{dx}-6\dfrac{dy}{dx}+8=0\;\Longrightarrow\;\dfrac{dy}{dx}=\dfrac{-8}{2y-6}=\dfrac{-4}{y-3}
Slope of the tangent at P, where y=-1.
m=\dfrac{-4}{-1-3}=1
Tangent by the point-slope form.
y+1=1\left(x-\tfrac14\right)\;\Longrightarrow\;4y+4=4x-1\;\Longrightarrow\;4x-4y-5=0
The normal has slope -1.
y+1=-\left(x-\tfrac14\right)\;\Longrightarrow\;4y+4=-4x+1\;\Longrightarrow\;4x+4y+3=0

The parabola is (y-3)^2=-8\left(x-\tfrac94\right), opening to the left. P is an end of its latus rectum.

\text{Tangent: }4x-4y-5=0,\qquad\text{Normal: }4x+4y+3=0

Reflection property of the parabola

Derivation Reflection property

Aim. Show that the tangent at any point P of a parabola makes equal angles with the focal line PF and with the line through P parallel to the axis.

A ray parallel to the axis strikes the parabola at P and is reflected through the focus F. Triangle FPT is isosceles, so the two marked angles at P are equal.

Take P(at^2,2at) on y^2=4ax. The tangent yy_1=2a(x+x_1) at P becomes
2at\,y=2a\left(x+at^2\right)\;\Longrightarrow\;ty=x+at^2
It meets the axis (y=0) at T.
T\left(-at^2,\,0\right)
Distance of T from the focus F(a,0), and the focal distance of P (x_1+a).
|FT|=a+at^2,\qquad|FP|=at^2+a
So |FT|=|FP| and triangle FPT is isosceles. Its base angles are equal.
\angle FPT=\angle FTP
The line through P parallel to the axis makes the angle \angle FTP with the tangent (corresponding angles with the axis TF).

Result\text{The tangent at P is equally inclined to PF and to the line through P parallel to the axis.}

By the law of reflection, a ray travelling parallel to the axis is reflected through the focus, and a ray from the focus is reflected parallel to the axis. This is why headlights, torches, satellite dishes and solar cookers are parabolic.

7.9, 7.9.1Ellipse: definition and standard equationOften examined

An ellipse is the set of all points in a plane for which the sum of the distances from two fixed points of the plane is the same (constant). The two fixed points are the foci (singular: focus) of the ellipse. The midpoint of the foci is the centre of the ellipse.

Derivation Standard equation of an ellipse

Aim. Find the equation of the ellipse with centre at the origin, foci F_1(-c,0) and F_2(c,0), and sum of focal distances 2a.

For every point P of the ellipse, |PF_1|+|PF_2|=2a, the length of the major axis.

Take any point P(x,y) of the ellipse. By the definition,
|PF_1|+|PF_2|=2a\;\Longrightarrow\;\sqrt{(x+c)^2+y^2}+\sqrt{(x-c)^2+y^2}=2a
Keep one root on each side, so squaring removes one of them.
\sqrt{(x+c)^2+y^2}=2a-\sqrt{(x-c)^2+y^2}
Square both sides.
(x+c)^2+y^2=4a^2-4a\sqrt{(x-c)^2+y^2}+(x-c)^2+y^2
Expand and cancel x^2, c^2 and y^2; then divide by 4.
4cx=4a^2-4a\sqrt{(x-c)^2+y^2}\;\Longrightarrow\;a\sqrt{(x-c)^2+y^2}=a^2-cx
Square again to remove the last root.
a^2\left(x^2-2cx+c^2+y^2\right)=a^4-2a^2cx+c^2x^2
Cancel -2a^2cx and collect terms.
\left(a^2-c^2\right)x^2+a^2y^2=a^2\left(a^2-c^2\right)
In triangle PF_1F_2 the sum of two sides is more than the third side, so 2a>2c, a>c and a^2-c^2>0. Call it b^2.
b^2=a^2-c^2\;\Longrightarrow\;b^2x^2+a^2y^2=a^2b^2
Divide both sides by a^2b^2.
\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1

Result\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1,\qquad b^2=a^2-c^2,\quad a>b>0

The ellipse is symmetric about both axes and lies inside -a\le x\le a, -b\le y\le b. The foci are always on the longer (major) axis.

If the foci are on the y-axis, F_1(0,-c) and F_2(0,c), the same steps give

\dfrac{x^2}{b^2}+\dfrac{y^2}{a^2}=1,\qquad a>b,\quad b^2=a^2-c^2

The larger denominator is always a^2, and it sits under the variable of the major axis.

7.9.1Elements of an ellipseOften examined

For the ellipse \dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1 with a>b and c^2=a^2-b^2:

Foci: The two fixed points F_1(-c,0) and F_2(c,0).
Major axis: The line through both foci: here the x-axis, y=0. The segment \overline{A_1A_2} of it inside the ellipse has length 2a (semi-major axis a).
Vertices: The points where the ellipse cuts its major axis: A_1(-a,0) and A_2(a,0).
Centre: The midpoint of the foci (and of the vertices): (0,0).
Minor axis: The line through the centre perpendicular to the major axis: here the y-axis, x=0. The segment inside the ellipse has length 2b (semi-minor axis b).
Co-vertices: The points where the ellipse cuts its minor axis: B_1(0,-b) and B_2(0,b).
Chord: A line segment with its end points on the ellipse.
Focal chord: A chord that passes through a focus, for example the major axis \overline{A_1A_2} or a latus rectum.
Latus rectum: A focal chord perpendicular to the major axis. There are two latera recta, one through each focus, each of length \dfrac{2b^2}{a}.
Eccentricity: e=\dfrac{c}{a}. Since c<a, 0<e<1.
Directrices: The two lines perpendicular to the major axis at distance \dfrac{a}{e} from the centre: x=\pm\dfrac{a}{e}=\pm\dfrac{a^2}{c}. They lie outside the ellipse.

Here a=5, b=3, c=4, e=\tfrac45. Vertices A_{1,2}(\pm5,0), co-vertices B_{1,2}(0,\mp3), foci F_{1,2}(\mp4,0), latera recta x=\pm4 of length \tfrac{2b^2}{a}=\tfrac{18}{5}, directrices x=\pm\tfrac{25}{4}.

Derivation Length of a latus rectum

Aim. Show that each latus rectum of \dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1 has length \dfrac{2b^2}{a}.

The latus rectum through F_2(c,0) is the line x=c. Put x=c in the equation.
\dfrac{y^2}{b^2}=1-\dfrac{c^2}{a^2}=\dfrac{a^2-c^2}{a^2}=\dfrac{b^2}{a^2}
Solve for y.
y^2=\dfrac{b^4}{a^2}\;\Longrightarrow\;y=\pm\dfrac{b^2}{a}
The ends are \left(c,\pm\dfrac{b^2}{a}\right), so the length is the difference of the y-coordinates.
l=\dfrac{b^2}{a}-\left(-\dfrac{b^2}{a}\right)=\dfrac{2b^2}{a}

Result\text{Length of each latus rectum}=\dfrac{2b^2}{a}

The same length \dfrac{2b^2}{a} holds for the hyperbola. For the parabola the latus rectum is 4a.

Eccentricity

From b^2=a^2-c^2, divide by a^2:

\dfrac{b^2}{a^2}=1-\dfrac{c^2}{a^2}=1-e^2\;\Longrightarrow\;e=\dfrac{c}{a}=\dfrac{\sqrt{a^2-b^2}}{a},\qquad b^2=a^2\left(1-e^2\right)

If e is close to 0, c is small and b\approx a: the ellipse is nearly a circle. If e is close to 1, b is small: the ellipse is long and thin. When e=0 the foci meet at the centre and the ellipse becomes a circle.

Derivation Focal distances and the focus-directrix property

Aim. Show that for P on the ellipse |PF_2|=a-ex, |PF_1|=a+ex, and that |PF_2|=e\,|PM| where M is on the directrix x=\dfrac{a}{e}.

Step 4 of the derivation of the ellipse gave a\sqrt{(x-c)^2+y^2}=a^2-cx. The root is |PF_2|. Divide by a and use c=ae.
|PF_2|=a-\dfrac{c}{a}x=a-ex
The sum of the focal distances is 2a.
|PF_1|=2a-|PF_2|=a+ex
Take e out as a common factor.
|PF_2|=e\left(\dfrac{a}{e}-x\right)
\dfrac{a}{e}-x is the distance from P to the line x=\dfrac{a}{e}.
|PF_2|=e\,|PM|

Result|PF_1|=a+ex,\quad |PF_2|=a-ex,\quad \dfrac{|PF|}{|PM|}=e<1

The directrix x=\dfrac{a}{e} really is the directrix belonging to the focus F_2: the ellipse is the conic with eccentricity e<1. The nearest and farthest points of the ellipse from a focus are the vertices, at distances a-c and a+c (useful for orbits).

Summary of the elements of an ellipse (a>b>0).
	\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1 (major axis on x-axis)	\dfrac{x^2}{b^2}+\dfrac{y^2}{a^2}=1 (major axis on y-axis)
Centre	(0,0)	(0,0)
Foci	(\pm c,0)	(0,\pm c)
Major axis	y=0, length 2a	x=0, length 2a
Minor axis	x=0, length 2b	y=0, length 2b
Vertices	(\pm a,0)	(0,\pm a)
Co-vertices	(0,\pm b)	(\pm b,0)
Directrices	x=\pm\dfrac{a}{e}	y=\pm\dfrac{a}{e}
Latus rectum	\dfrac{2b^2}{a}	\dfrac{2b^2}{a}
Eccentricity	e=\dfrac{c}{a}<1, c^2=a^2-b^2	e=\dfrac{c}{a}<1, c^2=a^2-b^2

Ellipse with centre at (h, k)

\dfrac{(x-h)^2}{a^2}+\dfrac{(y-k)^2}{b^2}=1\qquad\text{or}\qquad\dfrac{(x-h)^2}{b^2}+\dfrac{(y-k)^2}{a^2}=1\qquad(a>b)

For the first (major axis parallel to the x-axis): centre (h,k), foci (h\pm c,k), vertices (h\pm a,k), co-vertices (h,k\pm b), major axis y=k, minor axis x=h, directrices x=h\pm\dfrac{a}{e}. For the second, move along the y direction instead: foci (h,k\pm c), vertices (h,k\pm a), co-vertices (h\pm b,k), directrices y=k\pm\dfrac{a}{e}.

Example 18

Find the equation of an ellipse with foci (-3,0) and (3,0) and the sum of the distances of any point from the foci equal to 10.

Method 1 (from the definition, as in the book).

For P(x,y) on the ellipse, |PF_1|+|PF_2|=10. Isolate one root.
\sqrt{(x+3)^2+y^2}=10-\sqrt{(x-3)^2+y^2}
Square and simplify (x^2, y^2 and 9 cancel).
12x=100-20\sqrt{(x-3)^2+y^2}\;\Longrightarrow\;5\sqrt{(x-3)^2+y^2}=25-3x
Square again.
25\left(x^2-6x+9+y^2\right)=625-150x+9x^2\;\Longrightarrow\;16x^2+25y^2=400
Divide by 400.
\dfrac{x^2}{25}+\dfrac{y^2}{16}=1

Method 2 (using the elements). The centre is the midpoint (0,0), c=3 and 2a=10, so a=5 and b^2=a^2-c^2=25-9=16. The foci are on the x-axis, so the equation is \dfrac{x^2}{25}+\dfrac{y^2}{16}=1.

\dfrac{x^2}{25}+\dfrac{y^2}{16}=1

7.9.1Converting to standard form: elements and graph of an ellipseOften examined

An equation such as Ax^2+Cy^2+Dx+Ey+F=0 with A and C of the same sign is usually an ellipse. To find its elements:

Group the x terms and the y terms and take out the coefficients of x^2 and y^2.
Complete both squares and move the constant to the right side.
Divide by the right side to get \dfrac{(x-h)^2}{\ \cdot\ }+\dfrac{(y-k)^2}{\ \cdot\ }=1. The larger denominator is a^2.
Put X=x-h, Y=y-k, find every element in X,Y, then convert back with x=X+h, y=Y+k.

Example 19

Convert the equation of the ellipse 4x^2+9y^2+8x-36y+4=0 to standard form. Also find its elements and draw the graph.

Group and take out the coefficients.
4\left(x^2+2x\right)+9\left(y^2-4y\right)=-4
Complete the squares inside the brackets (add 1 and 4) and balance: 4\times1 and 9\times4 were added.
4(x+1)^2+9(y-2)^2=-4+4+36=36
Divide by 36.
\dfrac{(x+1)^2}{9}+\dfrac{(y-2)^2}{4}=1
Let X=x+1, Y=y-2. Then \dfrac{X^2}{3^2}+\dfrac{Y^2}{2^2}=1: the major axis is horizontal, with
a=3,\quad b=2,\quad c=\sqrt{a^2-b^2}=\sqrt5,\quad e=\dfrac{c}{a}=\dfrac{\sqrt5}{3}
Centre: X=0, Y=0.
x=-1,\ y=2\;\Longrightarrow\;(-1,2)
Foci: X=\pm\sqrt5, Y=0.
\left(-1-\sqrt5,\,2\right),\ \left(-1+\sqrt5,\,2\right)
Vertices: X=\pm3, Y=0.
(-4,2),\ (2,2)
Co-vertices: X=0, Y=\pm2.
(-1,0),\ (-1,4)
Major axis: Y=0, that is y=2 (length 2a=6). Minor axis: X=0, that is x=-1 (length 2b=4).
Directrices: X=\pm\dfrac{a^2}{c}=\pm\dfrac{9}{\sqrt5}.
x=-1\pm\dfrac{9}{\sqrt5},\ \text{that is } x=\dfrac{9-\sqrt5}{\sqrt5}\ \text{and}\ x=\dfrac{-9-\sqrt5}{\sqrt5}
Latus rectum: length \dfrac{2b^2}{a}=\dfrac{8}{3}.

Centre (-1,2), foci \left(-1\pm\sqrt5,\,2\right), vertices (-4,2) and (2,2), co-vertices (-1,0) and (-1,4), directrices x=-1\pm\tfrac{9}{\sqrt5}.

\dfrac{(x+1)^2}{9}+\dfrac{(y-2)^2}{4}=1

7.10, 7.10.1Line and ellipse: points of intersection and condition of tangencyOften examined

To find where a line meets an ellipse, substitute the line into the ellipse and solve the quadratic. Its discriminant decides the position: \Delta>0 two points, \Delta=0 one point (a tangent), \Delta<0 no point.

Example (points of intersection)

Find the points where the line x+2y=4 meets the ellipse \dfrac{x^2}{16}+\dfrac{y^2}{4}=1.

From the line, x=4-2y. Substitute in the ellipse.
\dfrac{(4-2y)^2}{16}+\dfrac{y^2}{4}=1
Expand: \dfrac{16-16y+4y^2}{16}=1-y+\dfrac{y^2}{4}.
1-y+\dfrac{y^2}{4}+\dfrac{y^2}{4}=1\;\Longrightarrow\;\dfrac{y^2}{2}-y=0
Factor.
y\left(\tfrac{y}{2}-1\right)=0\;\Longrightarrow\;y=0\ \text{or}\ y=2
Find x=4-2y.
y=0\Rightarrow x=4;\qquad y=2\Rightarrow x=0

(4,0)\ \text{and}\ (0,2)

Derivation Condition for y=mx+c to touch \dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1

Aim. Find the condition for the line y=mx+c to be a tangent to the ellipse \dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1.

Substitute y=mx+c and multiply by a^2b^2.
b^2x^2+a^2(mx+c)^2=a^2b^2
Arrange as a quadratic in x.
\left(b^2+a^2m^2\right)x^2+2a^2mc\,x+a^2\left(c^2-b^2\right)=0
For a tangent the discriminant must be zero.
4a^4m^2c^2-4\left(b^2+a^2m^2\right)a^2\left(c^2-b^2\right)=0
Divide by 4a^2 and expand; a^2m^2c^2 cancels.
-b^2c^2+b^4+a^2m^2b^2=0
Divide by b^2.
c^2=a^2m^2+b^2\;\Longrightarrow\;c=\pm\sqrt{a^2m^2+b^2}

Resultc^2=a^2m^2+b^2,\qquad y=mx\pm\sqrt{a^2m^2+b^2}

For every slope m an ellipse has two parallel tangents. The point of contact of y=mx+c is \left(-\dfrac{a^2m}{c},\ \dfrac{b^2}{c}\right). For an ellipse with centre (h,k) the condition becomes (c-k+mh)^2=a^2m^2+b^2.

Example 20

Find the equations of the tangents to the ellipse \dfrac{x^2}{2}+\dfrac{y^2}{16}=1 with slope \tfrac12.

Compare with \dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1 (the formula holds whichever denominator is larger).
a^2=2,\quad b^2=16,\quad m=\tfrac12
Condition of tangency.
c^2=a^2m^2+b^2=2\left(\tfrac14\right)+16=\tfrac{33}{2}\;\Longrightarrow\;c=\pm\sqrt{\tfrac{33}{2}}=\pm\dfrac{\sqrt{66}}{2}
Tangents y=mx+c; multiply by 2.
y=\tfrac12x\pm\dfrac{\sqrt{66}}{2}\;\Longrightarrow\;2y=x\pm\sqrt{66}
Check by substitution: y=\tfrac{x}{2}+c in 8x^2+y^2=16 gives \tfrac{33}{4}x^2+cx+c^2-16=0, whose discriminant c^2-33\left(c^2-16\right)=528-32c^2 is zero exactly when c^2=\tfrac{33}{2}.

The two tangents with slope \tfrac12 touch the ellipse at \left(-\tfrac{a^2m}{c},\tfrac{b^2}{c}\right) with c=\pm\tfrac{\sqrt{66}}{2}, that is near (-0.25,\,3.94) and (0.25,\,-3.94).

2y=x\pm\sqrt{66}

Example 21

Prove that the line \sqrt3x-2y+8=0 is a tangent to the ellipse \dfrac{x^2}{16}+\dfrac{y^2}{4}=1.

From the line, y=\dfrac{\sqrt3x+8}{2}. The ellipse is x^2+4y^2=16. Substitute.
x^2+4\cdot\dfrac{3x^2+16\sqrt3x+64}{4}=16
Simplify.
4x^2+16\sqrt3x+48=0\;\Longrightarrow\;x^2+4\sqrt3x+12=0
Discriminant.
\Delta=\left(4\sqrt3\right)^2-4(1)(12)=48-48=0
\Delta=0, so the line meets the ellipse at one point only: it is a tangent. The double root is x=-2\sqrt3, giving the point of contact.
\left(-2\sqrt3,\ 1\right)
Check with the condition: m=\tfrac{\sqrt3}{2}, c=4, a^2m^2+b^2=16\cdot\tfrac34+4=16=c^2.

\text{Tangent, touching at }\left(-2\sqrt3,\,1\right)

7.11, 7.11.1Tangent and normal to an ellipse at a pointOften examined

Derivation Tangent at P(x_1,y_1)

Aim. Find the tangent at the point P(x_1,y_1) of the ellipse \dfrac{(x-h)^2}{a^2}+\dfrac{(y-k)^2}{b^2}=1.

Clear the fractions: b^2(x-h)^2+a^2(y-k)^2=a^2b^2 (1). Differentiate with respect to x.
2b^2(x-h)+2a^2(y-k)\dfrac{dy}{dx}=0\;\Longrightarrow\;\dfrac{dy}{dx}=-\dfrac{b^2(x-h)}{a^2(y-k)}
Point-slope form at P(x_1,y_1); multiply by a^2(y_1-k).
a^2(y_1-k)(y-y_1)=-b^2(x_1-h)(x-x_1)
Write x-x_1=(x-h)-(x_1-h) and y-y_1=(y-k)-(y_1-k) and expand.
b^2(x_1-h)(x-h)+a^2(y_1-k)(y-k)=b^2(x_1-h)^2+a^2(y_1-k)^2
P lies on the ellipse, so by (1) the right side is a^2b^2. Divide by a^2b^2.
\dfrac{(x_1-h)(x-h)}{a^2}+\dfrac{(y_1-k)(y-k)}{b^2}=1

Result\dfrac{(x_1-h)(x-h)}{a^2}+\dfrac{(y_1-k)(y-k)}{b^2}=1;\qquad\dfrac{xx_1}{a^2}+\dfrac{yy_1}{b^2}=1\ \text{ for centre }(0,0)

Again the replacement rule works: replace x^2 by xx_1 and y^2 by yy_1 in the equation of the ellipse.

Derivation Normal at P(x_1,y_1)

Aim. Find the normal at P(x_1,y_1) of the same ellipse.

The slope of the normal is the negative reciprocal of the tangent's slope.
m_1=\dfrac{a^2(y_1-k)}{b^2(x_1-h)}
Point-slope form, multiplied by b^2(x_1-h).
b^2(x_1-h)(y-y_1)=a^2(y_1-k)(x-x_1)
Bring all terms to one side.
a^2(y_1-k)(x-x_1)-b^2(x_1-h)(y-y_1)=0

Result\begin{aligned}a^2(y_1-k)(x-x_1)-b^2(x_1-h)(y-y_1)&=0\\ a^2y_1(x-x_1)-b^2x_1(y-y_1)&=0\ \text{ for centre }(0,0)\end{aligned}

For the centre at the origin and x_1y_1\neq0, dividing by x_1y_1 gives the easy-to-remember form \dfrac{a^2x}{x_1}-\dfrac{b^2y}{y_1}=a^2-b^2.

Example 22

Find the equations of the tangent and normal to the ellipse x^2+2y^2-2x+4y=0 at the point P(0,0).

Method 1 (differentiation).

Differentiate with respect to x.
2x+4y\dfrac{dy}{dx}-2+4\dfrac{dy}{dx}=0\;\Longrightarrow\;\dfrac{dy}{dx}=-\dfrac{2x-2}{4y+4}=-\dfrac{x-1}{2y+2}
Slope at (0,0).
m=-\dfrac{0-1}{0+2}=\dfrac12
Tangent.
y-0=\tfrac12(x-0)\;\Longrightarrow\;x-2y=0
Normal (slope -2).
y=-2x\;\Longrightarrow\;2x+y=0

Method 2 (standard form).

Complete the squares.
(x-1)^2-1+2\left[(y+1)^2-1\right]=0\;\Longrightarrow\;\dfrac{(x-1)^2}{3}+\dfrac{(y+1)^2}{3/2}=1
So h=1, k=-1, a^2=3, b^2=\tfrac32 and (x_1,y_1)=(0,0). Tangent:
\dfrac{(0-1)(x-1)}{3}+\dfrac{(0+1)(y+1)}{3/2}=1\;\Longrightarrow\;-x+1+2y+2=3\;\Longrightarrow\;x-2y=0
Normal a^2(y_1-k)(x-x_1)-b^2(x_1-h)(y-y_1)=0:
3(1)(x)-\tfrac32(-1)(y)=0\;\Longrightarrow\;3x+\tfrac32y=0\;\Longrightarrow\;2x+y=0

\text{Tangent: }x-2y=0,\qquad\text{Normal: }2x+y=0

Reflection property of the ellipse

The tangent at any point P of an ellipse makes equal angles with the focal radii PF_1 and PF_2 (equivalently, the normal at P bisects the angle F_1PF_2). So a ray of light or a sound wave starting at one focus is reflected by the ellipse through the other focus (proved in Exercise 7.7 Q7).

A ray from one focus is reflected by the ellipse through the other focus: the tangent at P makes equal angles with PF_1 and PF_2.

7.12, 7.12.1Hyperbola: definition and standard equationOften examined

A hyperbola is the set of all points in a plane for which the difference of the distances from two fixed points of the plane is the same (constant). The two fixed points are the foci of the hyperbola, and their midpoint is its centre. We take the difference as a positive number, \big||PF_1|-|PF_2|\big|=2a; points nearer to F_2 form one branch and points nearer to F_1 form the other.

Derivation Standard equation of a hyperbola

Aim. Find the equation of the hyperbola with centre at the origin, foci F_1(-c,0) and F_2(c,0), and difference of focal distances 2a.

For every point P of the hyperbola, \big||PF_1|-|PF_2|\big|=2a, the length of the transverse axis. Here a=3, b=4, c=5.

Take any point P(x,y) of the hyperbola. By the definition,
\sqrt{(x+c)^2+y^2}-\sqrt{(x-c)^2+y^2}=\pm2a
Keep one root on each side.
\sqrt{(x+c)^2+y^2}=\pm2a+\sqrt{(x-c)^2+y^2}
Square both sides.
(x+c)^2+y^2=4a^2\pm4a\sqrt{(x-c)^2+y^2}+(x-c)^2+y^2
Cancel x^2, c^2, y^2 and divide by 4.
cx-a^2=\pm a\sqrt{(x-c)^2+y^2}
Square again.
c^2x^2-2a^2cx+a^4=a^2\left(x^2-2cx+c^2+y^2\right)
Cancel -2a^2cx and collect terms.
\left(c^2-a^2\right)x^2-a^2y^2=a^2\left(c^2-a^2\right)
In triangle PF_1F_2 the difference of two sides is less than the third side, so 2a<2c, c>a and c^2-a^2>0. Call it b^2.
b^2=c^2-a^2\;\Longrightarrow\;b^2x^2-a^2y^2=a^2b^2
Divide by a^2b^2.
\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1

Result\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1,\qquad b^2=c^2-a^2

The curve is symmetric about both axes. Since \dfrac{x^2}{a^2}=1+\dfrac{y^2}{b^2}\ge1, no point lies between x=-a and x=a: the hyperbola has two separate branches. Here a need not be larger than b.

With the foci on the y-axis, (0,\pm c), the same steps give

\dfrac{y^2}{a^2}-\dfrac{x^2}{b^2}=1,\qquad b^2=c^2-a^2

a^2 is always under the positive term, whichever denominator is larger.

When the y^2 term is positive the branches open up and down, the foci are on the y-axis and the asymptotes are y=\pm\tfrac{a}{b}x (here \pm\tfrac43x).

7.12.1Elements and asymptotes of a hyperbolaOften examined

For the hyperbola \dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1 with c^2=a^2+b^2:

Foci: The two fixed points F_1(-c,0) and F_2(c,0).
Centre: The midpoint of the foci: (0,0).
Focal axis (transverse or real axis): The line through both foci: here y=0. The segment between the vertices has length 2a.
Conjugate axis (imaginary axis): The line through the centre perpendicular to the focal axis: here x=0. Its length is taken as 2b.
Vertices: The points where the hyperbola cuts its focal axis: A_1(-a,0) and A_2(a,0).
Co-vertices: The points B_1(0,-b) and B_2(0,b) on the conjugate axis. They are not on the curve.
Focal length: The distance between the foci, 2c.
Chord: A line segment with both end points on the hyperbola (the book takes both ends on the same branch).
Focal chord: A chord through a focus.
Latus rectum: The focal chord perpendicular to the focal axis. There are two, each of length \dfrac{2b^2}{a}.
Eccentricity: e=\dfrac{c}{a}. Since c>a, e>1. Also e^2=1+\dfrac{b^2}{a^2} and b^2=a^2\left(e^2-1\right).
Directrices: The lines x=\pm\dfrac{a}{e}=\pm\dfrac{a^2}{c}, perpendicular to the focal axis, between the two branches.

The asymptotes y=\pm\tfrac43x are the diagonals of the rectangle through the vertices (\pm3,0) and co-vertices (0,\pm4). The directrices x=\pm\tfrac{a}{e}=\pm\tfrac95 lie between the branches.

Derivation Asymptotes of the hyperbola

Aim. Show that the branches of \dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1 approach the lines y=\pm\dfrac{b}{a}x.

Solve the equation for y.
y=\pm\dfrac{b}{a}\sqrt{x^2-a^2}
Compare the curve with the line y=\dfrac{b}{a}x in the first quadrant. The vertical gap between them is
\dfrac{b}{a}\left(x-\sqrt{x^2-a^2}\right)
Multiply and divide by x+\sqrt{x^2-a^2} (rationalising).
\dfrac{b}{a}\cdot\dfrac{x^2-\left(x^2-a^2\right)}{x+\sqrt{x^2-a^2}}=\dfrac{ab}{x+\sqrt{x^2-a^2}}
As x\to\infty the denominator grows without limit, so the gap tends to 0. By symmetry the same happens in the other quadrants.
\lim_{x\to\infty}\dfrac{ab}{x+\sqrt{x^2-a^2}}=0

Result\text{Asymptotes: }y=\pm\dfrac{b}{a}x\quad\Longleftrightarrow\quad\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=0

The branches get closer and closer to these two lines but never meet them. The asymptotes are the diagonals of the rectangle with sides 2a and 2b centred at the centre, which makes the hyperbola easy to sketch. For \dfrac{y^2}{a^2}-\dfrac{x^2}{b^2}=1 the asymptotes are y=\pm\dfrac{a}{b}x.

If a=b, the hyperbola x^2-y^2=a^2 is called a rectangular (equilateral) hyperbola: its asymptotes y=\pm x are perpendicular and e=\sqrt2.

Derivation Focal distances and the focus-directrix property

Aim. Show that the hyperbola is the conic with eccentricity e>1: |PF_2|=e\,|PM|, M on the directrix x=\dfrac{a}{e}.

Step 4 of the derivation gave a\sqrt{(x-c)^2+y^2}=\pm\left(cx-a^2\right), and the root is |PF_2|. Divide by a and use c=ae.
|PF_2|=\left|ex-a\right|
Take out e.
|PF_2|=e\left|x-\dfrac{a}{e}\right|
\left|x-\dfrac{a}{e}\right| is the distance from P to the line x=\dfrac{a}{e}.
\dfrac{|PF_2|}{|PM|}=e

Result|PF_1|=|ex+a|,\quad|PF_2|=|ex-a|,\quad\dfrac{|PF|}{|PM|}=e>1

The directrix x=\dfrac{a}{e} goes with the focus (c,0) and x=-\dfrac{a}{e} with (-c,0). Example 28 uses this property to find a hyperbola from a focus, a directrix and e.

Summary of the elements of a hyperbola (corrected).
	\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1 (opens left-right)	\dfrac{y^2}{a^2}-\dfrac{x^2}{b^2}=1 (opens up-down)
Centre	(0,0)	(0,0)
Foci	(\pm c,0)	(0,\pm c)
Vertices	A_{1,2}(\mp a,0)	A_{1,2}(0,\mp a)
Co-vertices	B_{1,2}(0,\mp b)	B_{1,2}(\mp b,0)
Focal axis	y=0, length 2a	x=0, length 2a
Conjugate axis	x=0, length 2b	y=0, length 2b
Eccentricity	e=\dfrac{c}{a}>1, c^2=a^2+b^2	e=\dfrac{c}{a}>1, c^2=a^2+b^2
Directrices	x=\pm\dfrac{a}{e}	y=\pm\dfrac{a}{e}
Latus rectum	\dfrac{2b^2}{a}	\dfrac{2b^2}{a}
Asymptotes	y=\pm\dfrac{b}{a}x	y=\pm\dfrac{a}{b}x

Hyperbola with centre at (h, k)

\dfrac{(x-h)^2}{a^2}-\dfrac{(y-k)^2}{b^2}=1\qquad\text{or}\qquad\dfrac{(y-k)^2}{a^2}-\dfrac{(x-h)^2}{b^2}=1

For the first: centre (h,k), foci (h\pm c,k), vertices (h\pm a,k), co-vertices (h,k\pm b), focal axis y=k, conjugate axis x=h, directrices x=h\pm\dfrac{a}{e}, asymptotes y-k=\pm\dfrac{b}{a}(x-h). For the second: foci (h,k\pm c), vertices (h,k\pm a), directrices y=k\pm\dfrac{a}{e}, asymptotes y-k=\pm\dfrac{a}{b}(x-h).

7.12.1Finding the elements and the equation of a hyperbolaOften examined

The methods are the same as for the ellipse: complete the squares to reach a standard form, or use the given elements to find a, b, c and e and substitute. Remember c^2=a^2+b^2 and e=\dfrac{c}{a}.

Example 23

Convert the equation of the hyperbola 4x^2-9y^2-16x-18y-29=0 into standard form and find its elements.

Group and take out the coefficients.
4\left(x^2-4x\right)-9\left(y^2+2y\right)=29
Complete the squares and balance.
4\left[(x-2)^2-4\right]-9\left[(y+1)^2-1\right]=29\;\Longrightarrow\;4(x-2)^2-9(y+1)^2=36
Divide by 36.
\dfrac{(x-2)^2}{9}-\dfrac{(y+1)^2}{4}=1
Let X=x-2, Y=y+1: \dfrac{X^2}{3^2}-\dfrac{Y^2}{2^2}=1, so
a=3,\quad b=2,\quad c=\sqrt{a^2+b^2}=\sqrt{13}
Centre: X=Y=0.
(2,-1)
Vertices: X=\pm3, Y=0.
x-2=\pm3\;\Longrightarrow\;(-1,-1),\ (5,-1)
Co-vertices: X=0, Y=\pm2.
(2,-3),\ (2,1)
Foci: X=\pm\sqrt{13}, Y=0.
\left(2\pm\sqrt{13},\,-1\right)
Eccentricity:
e=\dfrac{c}{a}=\dfrac{\sqrt{13}}{3}
Focal axis: Y=0, that is y+1=0, length 2a=6. Conjugate axis: X=0, that is x-2=0, length 2b=4.
Directrices: X=\pm\dfrac{a}{e}=\pm\dfrac{a^2}{c}=\pm\dfrac{9}{\sqrt{13}}.
x=2\pm\dfrac{9}{\sqrt{13}}
Latus rectum:
\dfrac{2b^2}{a}=\dfrac{2(4)}{3}=\dfrac83
Asymptotes: Y=\pm\dfrac{b}{a}X.
y+1=\pm\tfrac23(x-2)

Centre (2,-1), vertices (-1,-1) and (5,-1), foci \left(2\pm\sqrt{13},-1\right), directrices x=2\pm\tfrac{9}{\sqrt{13}}, asymptotes y+1=\pm\tfrac23(x-2) (dashed).

\dfrac{(x-2)^2}{9}-\dfrac{(y+1)^2}{4}=1

Example 24

Find the equation of the hyperbola with centre at (2,4) whose semi-transverse and semi-conjugate axes are 3 and 5 units respectively, the transverse axis being parallel to the x-axis.

The transverse axis is parallel to the x-axis, so use \dfrac{(x-h)^2}{a^2}-\dfrac{(y-k)^2}{b^2}=1 with
h=2,\quad k=4,\quad a=3,\quad b=5
Substitute.
\dfrac{(x-2)^2}{9}-\dfrac{(y-4)^2}{25}=1

\dfrac{(x-2)^2}{9}-\dfrac{(y-4)^2}{25}=1

Example 25

Find the equation of the hyperbola with centre at (0,2) that passes through the points (1,1) and (4,7) (transverse axis horizontal).

With centre (0,2) the equation is
\dfrac{x^2}{a^2}-\dfrac{(y-2)^2}{b^2}=1\qquad(1)
It passes through (1,1).
\dfrac{1}{a^2}-\dfrac{1}{b^2}=1\qquad(2)
It passes through (4,7).
\dfrac{16}{a^2}-\dfrac{25}{b^2}=1\qquad(3)
Multiply (2) by 16 and subtract (3).
\dfrac{9}{b^2}=15\;\Longrightarrow\;b^2=\dfrac{9}{15}=\dfrac35
Put this in (2).
\dfrac{1}{a^2}=1+\dfrac{15}{9}=\dfrac{24}{9}\;\Longrightarrow\;a^2=\dfrac{9}{24}=\dfrac38
Substitute in (1) and clear fractions.
\dfrac{x^2}{3/8}-\dfrac{(y-2)^2}{3/5}=1\;\Longrightarrow\;8x^2-5(y-2)^2=3
Check: (1,1) gives 8-5=3 and (4,7) gives 128-125=3.

8x^2-5(y-2)^2=3\quad\text{or}\quad 8x^2-5y^2+20y-23=0

Example 26

Find the equation of the hyperbola with centre at the origin, eccentricity \tfrac32 and length of latus rectum 3.

Take \dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1. Given e=\tfrac32 and \dfrac{2b^2}{a}=3.
Express b^2 in terms of a using c=ae=\tfrac{3a}{2}.
b^2=c^2-a^2=\dfrac{9a^2}{4}-a^2=\dfrac{5a^2}{4}
Put this in the latus rectum condition.
\dfrac{2}{a}\cdot\dfrac{5a^2}{4}=3\;\Longrightarrow\;\dfrac{5a}{2}=3\;\Longrightarrow\;a=\dfrac65,\ a^2=\dfrac{36}{25}
Then
b^2=\dfrac54\cdot\dfrac{36}{25}=\dfrac95
Substitute.
\dfrac{x^2}{36/25}-\dfrac{y^2}{9/5}=1\;\Longrightarrow\;\dfrac{25x^2}{36}-\dfrac{5y^2}{9}=1

\dfrac{25x^2}{36}-\dfrac{5y^2}{9}=1

Example 27

Find the equation of the hyperbola in standard form with centre at the origin and eccentricity \tfrac43, if the length of its semi-conjugate axis is 3 and the conjugate axis is along the y-axis.

The conjugate axis is the y-axis, so the focal axis is the x-axis: \dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1 with b=3.
Use b^2=a^2\left(e^2-1\right).
9=a^2\left(\dfrac{16}{9}-1\right)=\dfrac{7a^2}{9}\;\Longrightarrow\;a^2=\dfrac{81}{7}
Substitute.
\dfrac{x^2}{81/7}-\dfrac{y^2}{9}=1\;\Longrightarrow\;\dfrac{7x^2}{81}-\dfrac{y^2}{9}=1

\dfrac{7x^2}{81}-\dfrac{y^2}{9}=1

Example 28

Find the equation of the hyperbola with centre at the origin and eccentricity \sqrt3, one of whose foci is (3,0) and the corresponding directrix is x=1.

Method 1 (focus-directrix property).

For P(x,y) on the hyperbola, |PF|=e\,|PM| with F(3,0) and the directrix x-1=0.
\sqrt{(x-3)^2+y^2}=\sqrt3\,|x-1|
Square both sides.
x^2-6x+9+y^2=3\left(x^2-2x+1\right)
Simplify.
2x^2-y^2=6\;\Longrightarrow\;\dfrac{x^2}{3}-\dfrac{y^2}{6}=1

Method 2 (elements). The focus (3,0) gives c=ae=3, so a\sqrt3=3, a=\sqrt3, a^2=3 and b^2=c^2-a^2=9-3=6. Check: the directrix x=\dfrac{a}{e}=\dfrac{\sqrt3}{\sqrt3}=1, as given.

\dfrac{x^2}{3}-\dfrac{y^2}{6}=1

Example (line and hyperbola)

Find the points where the line y=x-3 meets the hyperbola \dfrac{x^2}{9}-\dfrac{y^2}{4}=1.

Substitute y=x-3 and multiply by 36.
4x^2-9(x-3)^2=36\;\Longrightarrow\;-5x^2+54x-117=0\;\Longrightarrow\;5x^2-54x+117=0
Solve: \Delta=54^2-4(5)(117)=2916-2340=576=24^2.
x=\dfrac{54\pm24}{10}=3\ \text{or}\ \dfrac{39}{5}
Find y=x-3.
(3,0)\ \text{and}\ \left(\tfrac{39}{5},\tfrac{24}{5}\right)

(3,0)\ \text{and}\ \left(\tfrac{39}{5},\tfrac{24}{5}\right)

7.13Line and hyperbola: condition of tangencyOften examined

As with the ellipse, substitute the line into the hyperbola. The quadratic in x has two distinct roots (two points), equal roots (a tangent) or no real roots (no common point).

Derivation Condition for y=mx+c to touch \dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1

Aim. Find the condition for the line y=mx+c to be a tangent to \dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1.

Substitute y=mx+c into b^2x^2-a^2y^2=a^2b^2.
b^2x^2-a^2(mx+c)^2=a^2b^2
Arrange as a quadratic in x.
\left(b^2-a^2m^2\right)x^2-2a^2mc\,x-a^2\left(c^2+b^2\right)=0
For a tangent the discriminant is zero.
4a^4m^2c^2+4\left(b^2-a^2m^2\right)a^2\left(c^2+b^2\right)=0
Divide by 4a^2 and expand; a^2m^2c^2 cancels.
b^2c^2+b^4-a^2m^2b^2=0
Divide by b^2.
c^2=a^2m^2-b^2\;\Longrightarrow\;c=\pm\sqrt{a^2m^2-b^2}

Resultc^2=a^2m^2-b^2,\qquad y=mx\pm\sqrt{a^2m^2-b^2}

Tangents with slope m exist only when a^2m^2>b^2, that is |m|>\dfrac{b}{a}: every tangent is steeper than the asymptotes. The point of contact is \left(-\dfrac{a^2m}{c},\ -\dfrac{b^2}{c}\right). For centre (h,k) the condition is (c-k+mh)^2=a^2m^2-b^2.

Example 29

Find the equations of the tangents to the hyperbola \dfrac{x^2}{3}-\dfrac{y^2}{2}=1 with slope \tfrac73.

Here a^2=3, b^2=2, m=\tfrac73. Condition of tangency:
c^2=a^2m^2-b^2=3\left(\tfrac{49}{9}\right)-2=\tfrac{49}{3}-2=\tfrac{43}{3}
So c=\pm\sqrt{\tfrac{43}{3}}=\pm\dfrac{\sqrt{129}}{3}.
y=\tfrac73x\pm\dfrac{\sqrt{129}}{3}\;\Longrightarrow\;3y=7x\pm\sqrt{129}

3y=7x\pm\sqrt{129}

Example 30

Check whether the line x-y-1=0 is a tangent to the hyperbola \dfrac{x^2}{4}-\dfrac{y^2}{3}=1.

The hyperbola is 3x^2-4y^2=12. From the line, y=x-1. Substitute.
3x^2-4(x-1)^2=12\;\Longrightarrow\;-x^2+8x-16=0\;\Longrightarrow\;x^2-8x+16=0
Discriminant.
\Delta=(-8)^2-4(1)(16)=64-64=0
\Delta=0, so the line is a tangent. The double root x=4 gives y=3: the point of contact is (4,3).
Check with the condition: m=1, c=-1, a^2m^2-b^2=4-3=1=c^2.

The line touches the right branch at (4,3). The dashed lines are the asymptotes y=\pm\tfrac{\sqrt3}{2}x.

\text{Tangent at }(4,3)

7.14Tangent and normal to a hyperbola at a pointOften examined

Derivation Tangent at P(x_1,y_1)

Aim. Find the tangent at the point P(x_1,y_1) of \dfrac{(x-h)^2}{a^2}-\dfrac{(y-k)^2}{b^2}=1.

Write the curve as b^2(x-h)^2-a^2(y-k)^2=a^2b^2 (1) and differentiate.
2b^2(x-h)-2a^2(y-k)\dfrac{dy}{dx}=0\;\Longrightarrow\;\dfrac{dy}{dx}=\dfrac{b^2(x-h)}{a^2(y-k)}
Point-slope form at P, multiplied by a^2(y_1-k).
a^2(y_1-k)(y-y_1)=b^2(x_1-h)(x-x_1)
Write x-x_1=(x-h)-(x_1-h), y-y_1=(y-k)-(y_1-k) and rearrange.
b^2(x_1-h)(x-h)-a^2(y_1-k)(y-k)=b^2(x_1-h)^2-a^2(y_1-k)^2
P is on the curve, so by (1) the right side is a^2b^2. Divide by a^2b^2.
\dfrac{(x_1-h)(x-h)}{a^2}-\dfrac{(y_1-k)(y-k)}{b^2}=1

Result\dfrac{(x_1-h)(x-h)}{a^2}-\dfrac{(y_1-k)(y-k)}{b^2}=1;\qquad\dfrac{xx_1}{a^2}-\dfrac{yy_1}{b^2}=1\ \text{ for centre }(0,0)

The book's longer form b^2(x_1-h)x-a^2(y_1-k)y+a^2k(y_1-k)-b^2h(x_1-h)-a^2b^2=0 is the same line. For the centre at the origin it becomes b^2x_1x-a^2y_1y=a^2b^2.

Derivation Normal at P(x_1,y_1)

Aim. Find the normal at P(x_1,y_1) of the same hyperbola.

The slope of the normal is the negative reciprocal of the tangent's slope.
m_1=-\dfrac{a^2(y_1-k)}{b^2(x_1-h)}
Point-slope form, multiplied by b^2(x_1-h).
b^2(x_1-h)(y-y_1)=-a^2(y_1-k)(x-x_1)
Collect terms.
a^2(y_1-k)(x-x_1)+b^2(x_1-h)(y-y_1)=0

Result\begin{aligned}a^2(y_1-k)(x-x_1)+b^2(x_1-h)(y-y_1)&=0\\ a^2y_1x+b^2x_1y-\left(a^2+b^2\right)x_1y_1&=0\ \text{ for centre }(0,0)\end{aligned}

For the centre at the origin and x_1y_1\neq0 this can also be written \dfrac{a^2x}{x_1}+\dfrac{b^2y}{y_1}=a^2+b^2.

Example 31

Find the equations of the tangent and normal to the hyperbola 4x^2-9y^2-16x-18y-29=0 at the point (5,-1).

Check: 100-9-80+18-29=0, so the point is on the curve. Differentiate.
8x-18y\dfrac{dy}{dx}-16-18\dfrac{dy}{dx}=0\;\Longrightarrow\;\dfrac{dy}{dx}=\dfrac{8(x-2)}{18(y+1)}=\dfrac{4(x-2)}{9(y+1)}
At (5,-1) the denominator is zero: the slope does not exist, so the tangent is vertical.
x=5\quad\text{or}\quad x-5=0
The normal is perpendicular to a vertical line, so it is horizontal through (5,-1).
y=-1\quad\text{or}\quad y+1=0
Check with the formula: the curve is \dfrac{(x-2)^2}{9}-\dfrac{(y+1)^2}{4}=1 (Example 23) and (5,-1) is its vertex. Tangent: \dfrac{3(x-2)}{9}-0=1\Rightarrow x=5.

At the vertex (5,-1) the tangent is the vertical line x=5 and the normal is the focal axis y=-1.

\text{Tangent: }x-5=0,\qquad\text{Normal: }y+1=0

Reflection property of the hyperbola

The tangent at any point P of a hyperbola bisects the angle F_1PF_2 between the focal radii (proved in Exercise 7.9 Q7). So a ray of light aimed at one focus and reflected by the outer side of a hyperbolic mirror travels towards the other focus. Reflecting telescopes (the Cassegrain design) combine a parabolic and a hyperbolic mirror in this way.

Tangents and normals of the four standard conics.
Conic	Tangent at (x_1,y_1)	Normal at (x_1,y_1)	Tangent with slope m
x^2+y^2=r^2	xx_1+yy_1=r^2	x_1y-y_1x=0	y=mx\pm r\sqrt{1+m^2}
y^2=4ax	yy_1=2a(x+x_1)	y_1(x-x_1)+2a(y-y_1)=0	y=mx+\dfrac{a}{m}
\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1	\dfrac{xx_1}{a^2}+\dfrac{yy_1}{b^2}=1	a^2y_1(x-x_1)-b^2x_1(y-y_1)=0	y=mx\pm\sqrt{a^2m^2+b^2}
\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1	\dfrac{xx_1}{a^2}-\dfrac{yy_1}{b^2}=1	a^2y_1(x-x_1)+b^2x_1(y-y_1)=0	y=mx\pm\sqrt{a^2m^2-b^2}

7.15Application of conic sectionsOften examined

Conic sections appear in astronomy, optics, engineering and architecture. In a word problem, first choose axes so that the conic has a standard equation (usually with its centre or vertex at the origin), then use a known point to find the unknown constant, and finally substitute the point asked about.

Example 32

A train track is 8 ft wide and goes through a semicircular tunnel of radius 13 ft. How high is the tunnel at the edge of the track?

Put the centre of the tunnel at the origin. The tunnel is the upper half of the circle
x^2+y^2=13^2=169\qquad(1)
The track is 8 ft wide and centred, so its edges are at x=\pm4. Let the height there be h: the points A(-4,h) and B(4,h) are on the circle.
Substitute B(4,h) in (1).
16+h^2=169\;\Longrightarrow\;h^2=153\;\Longrightarrow\;h=\sqrt{153}=3\sqrt{17}\approx12.4\,\mathrm{ft}

Semicircular tunnel of radius 13 ft with the 8 ft track centred under it.

h=\sqrt{153}\approx12.4\,\mathrm{ft}

Example 33

The cable of a suspension bridge hangs in the shape of a parabola. The towers supporting the cable are 200 ft apart and 100 ft high. If the lowest point of the cable is 30 ft above the bridge at its midpoint, how high is the cable at a horizontal distance of 40 ft from either tower?

Take the origin on the road midway between the towers. The vertex is V(0,30) and the parabola opens upward.
x^2=4a(y-30)\qquad(1)
The top of a tower, B(100,100), is on the cable.
100^2=4a(100-30)\;\Longrightarrow\;4a=\dfrac{10000}{70}=\dfrac{1000}{7}
So the cable is
x^2=\dfrac{1000}{7}(y-30)\qquad(2)
40 ft from a tower is 100-40=60 ft from the middle. Put x=60, y=h in (2).
3600=\dfrac{1000}{7}(h-30)\;\Longrightarrow\;h-30=\dfrac{3600\times7}{1000}=25.2
Hence
h=55.2\,\mathrm{ft}

Towers 200 ft apart and 100 ft high; with the lowest point 30 ft above the road (the value the book's solution uses).

h=55.2\,\mathrm{ft}

7.16Use of conic sectionsOften examined

Each conic has a property that makes it useful: a circle keeps a fixed distance from its centre, a parabola reflects rays parallel to its axis through its focus, an ellipse reflects rays from one focus to the other, and a hyperbola keeps a constant difference of distances from its foci.

Example 34

The headlight of an automobile is in the shape of a parabola. The bulb is placed at the focus, which is 2 inches from its vertex. The depth of the headlight is 3 inches. What is the width of the headlight at its opening?

Put the vertex at the origin with the parabola opening to the right. The focus is (a,0)=(2,0), so a=2.
y^2=4ax=8x\qquad(1)
Let the width be w. The rim is 3 in from the vertex, so its ends are A\left(3,-\tfrac{w}{2}\right) and B\left(3,\tfrac{w}{2}\right). Put B in (1).
\left(\dfrac{w}{2}\right)^2=8(3)=24
Take the positive square root.
\dfrac{w}{2}=\sqrt{24}=2\sqrt6\;\Longrightarrow\;w=4\sqrt6\approx9.8\,\text{inches}

The bulb is at the focus, 2 in from the vertex; the reflector is 3 in deep.

w=4\sqrt6\approx9.8\,\text{inches}

Example 35

According to Kepler's law, planets have elliptical orbits with the Sun at one focus. If the longest distance of a planet from the Sun is 7.4 billion km and the shortest is 4.4 billion km, write the equation of the orbit taking its centre at the origin.

Take the major axis along the x-axis: \dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1 (1). Put the Sun at the focus F_1. The planet is farthest at the vertex A_2 and nearest at A_1.
Farthest distance |F_1A_2|=a+c and nearest distance |F_1A_1|=a-c.
a+c=7.4\quad(2),\qquad a-c=4.4\quad(3)
Add and subtract (2) and (3).
2a=11.8\Rightarrow a=5.9,\qquad 2c=3\Rightarrow c=1.5
Find b^2=a^2-c^2.
b^2=5.9^2-1.5^2=34.81-2.25=32.56,\qquad b\approx5.7
Substitute in (1) (distances in billions of km).
\dfrac{x^2}{34.81}+\dfrac{y^2}{32.56}=1\quad\text{or}\quad\dfrac{x^2}{(5.9)^2}+\dfrac{y^2}{(5.7)^2}=1

Distances in billions of km. Nearest point A_1: a-c=4.4; farthest point A_2: a+c=7.4. The orbit is nearly circular because e=\tfrac{c}{a}\approx0.25.

\dfrac{x^2}{34.81}+\dfrac{y^2}{32.56}=1

Example 36

The cross section of a nuclear cooling tower is in the shape of the hyperbola \dfrac{x^2}{30^2}-\dfrac{y^2}{44^2}=1. The tower is 150 m tall and the distance from the top of the tower to the centre of the hyperbola is half the distance from the base of the tower to the centre. Find the diameters of the top and the base of the tower.

Let A and B be the centres of the top and the base, on the y-axis. Given |OA|=\tfrac12|OB| and |OA|+|OB|=150.
|OA|+2|OA|=150\;\Longrightarrow\;|OA|=50,\quad|OB|=100
Let r_1, r_2 be the radii of the top and base. The points P(r_1,50) and Q(r_2,-100) are on the hyperbola. For P:
\dfrac{r_1^2}{900}=1+\left(\dfrac{50}{44}\right)^2=1+\left(\dfrac{25}{22}\right)^2=\dfrac{22^2+25^2}{22^2}
Take the square root.
r_1=30\cdot\dfrac{\sqrt{1109}}{22}\approx45.41\,\mathrm{m}\;\Longrightarrow\;\text{top diameter}=2r_1\approx90.82\,\mathrm{m}
For Q:
\dfrac{r_2^2}{900}=1+\left(\dfrac{100}{44}\right)^2=\dfrac{11^2+25^2}{11^2}\;\Longrightarrow\;r_2=30\cdot\dfrac{\sqrt{746}}{11}\approx74.49\,\mathrm{m}
So the base diameter is
2r_2\approx148.98\,\mathrm{m}

Height 150 m: the top is 50 m above the centre O and the base 100 m below it.

\text{Top}\approx90.8\,\mathrm{m},\qquad\text{base}\approx149.0\,\mathrm{m}

Where each conic is used

Conic	Property used	Applications
Circle	Every point is at the same distance from the centre	Wheels and gears, tunnels and arches, the coverage area of a mobile phone tower or radio transmitter (Exercise 7.10 Q1), circular motion
Parabola	Rays parallel to the axis reflect through the focus (and back)	Car headlights, torches (flashlights) and searchlights with the bulb at the focus; satellite dishes, radio telescopes and parabolic microphones with the receiver at the focus; solar cookers; cables of suspension bridges; paths of projectiles and thrown balls
Ellipse	Rays from one focus reflect through the other focus; the sum of focal distances is constant	Orbits of planets and satellites with the Sun or Earth at a focus (Kepler's first law); elliptic paths of electrons round the nucleus in the Bohr-Sommerfeld model of the atom; whispering galleries; lithotripsy; elliptical gears and arches
Hyperbola	The difference of distances from the foci is constant; rays aimed at one focus reflect towards the other	Radio navigation (LORAN): two stations send signals at the same time, and the constant time difference places a ship on a hyperbola; hyperbolic mirrors in reflecting telescopes; nuclear cooling towers and other hyperbolic structures in architecture; the path of a comet that passes the Sun only once

Summary: the four conics comparedOften examined

This table collects the standard forms (centre or vertex at the origin) of the four conics. For a centre or vertex at (h,k), replace x by x-h and y by y-k everywhere.

The four conics side by side.
	Circle	Parabola	Ellipse	Hyperbola
Definition (locus of P)	\|PC\|=r	\|PF\|=\|PM\|	\|PF_1\|+\|PF_2\|=2a	\big\|\|PF_1\|-\|PF_2\|\big\|=2a
Standard equation	x^2+y^2=r^2	y^2=4ax	\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1, a>b	\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1
General form	x^2+y^2+2gx+2fy+c=0	y=ax^2+bx+c or x=ay^2+by+c	Ax^2+Cy^2+Dx+Ey+F=0, AC>0	Ax^2+Cy^2+Dx+Ey+F=0, AC<0
Eccentricity	e=0	e=1	e=\dfrac{c}{a}<1	e=\dfrac{c}{a}>1
Relation	(radius r)	(focal length a)	c^2=a^2-b^2	c^2=a^2+b^2
Focus / foci	centre (0,0)	(a,0)	(\pm c,0)	(\pm c,0)
Vertices	none	(0,0)	(\pm a,0)	(\pm a,0)
Directrix	none	x=-a	x=\pm\dfrac{a}{e}	x=\pm\dfrac{a}{e}
Axes	every diameter is an axis of symmetry	axis y=0	major 2a, minor 2b	transverse 2a, conjugate 2b
Latus rectum	the diameter 2r	4a	\dfrac{2b^2}{a}	\dfrac{2b^2}{a}
Asymptotes	none	none	none	y=\pm\dfrac{b}{a}x
Tangent at (x_1,y_1)	xx_1+yy_1=r^2	yy_1=2a(x+x_1)	\dfrac{xx_1}{a^2}+\dfrac{yy_1}{b^2}=1	\dfrac{xx_1}{a^2}-\dfrac{yy_1}{b^2}=1
Condition for y=mx+c to touch	c^2=r^2\left(1+m^2\right)	c=\dfrac{a}{m}	c^2=a^2m^2+b^2	c^2=a^2m^2-b^2
Reflection property	a ray through the centre returns along itself	rays parallel to the axis pass through the focus	rays from one focus pass through the other	rays aimed at one focus go towards the other

Formula sheet

Eccentricity (definition of a conic)

e=\dfrac{|PF|}{|PM|}

P: point on the conic, F: focus, |PM|: distance of P from the directrix. e=0 circle, 0<e<1 ellipse, e=1 parabola, e>1 hyperbola

Circle, standard form

(x-h)^2+(y-k)^2=r^2

(h,k): centre, r: radius. With centre at the origin: x^2+y^2=r^2

Circle, general form

x^2+y^2+2gx+2fy+c=0

Centre (-g,-f), radius r=\sqrt{g^2+f^2-c}. Real circle if g^2+f^2-c>0, point circle if =0, imaginary if <0

Distance of a point from a line

d=\dfrac{|ax_1+by_1+c|}{\sqrt{a^2+b^2}}

(x_1,y_1): the point, ax+by+c=0: the line. Used to compare the distance of the centre from a line with r

Line and circle

d<r:\ \text{two points};\quad d=r:\ \text{tangent};\quad d>r:\ \text{no point}

d: distance of the centre from the line, r: radius. Equivalently the discriminant of the quadratic after substitution is >0, =0, <0

Condition of tangency to x^2+y^2=r^2

c^2=r^2\left(1+m^2\right),\qquad y=mx\pm r\sqrt{1+m^2}

y=mx+c: the line, m: slope, r: radius. The second form gives the two tangents of slope m

Tangent to a circle at (x_1,y_1)

xx_1+yy_1+g(x+x_1)+f(y+y_1)+c=0

(x_1,y_1): point on the circle x^2+y^2+2gx+2fy+c=0. For x^2+y^2=r^2: xx_1+yy_1=r^2

Normal to a circle at (x_1,y_1)

(y_1+f)(x-x_1)-(x_1+g)(y-y_1)=0

The normal is the line through (x_1,y_1) and the centre (-g,-f)

Position of a point

S_1=x_1^2+y_1^2+2gx_1+2fy_1+c\ \ \begin{cases}>0 & \text{outside}\\=0 & \text{on}\\<0 & \text{inside}\end{cases}

S_1: the circle expression evaluated at (x_1,y_1), after making the coefficients of x^2, y^2 equal to 1

Length of tangent from an external point

|PA|=\sqrt{x_1^2+y_1^2+2gx_1+2fy_1+c}=\sqrt{|PC|^2-r^2}

P(x_1,y_1): external point, A: point of contact, C: centre

Parabola, standard forms

y^2=\pm4ax,\qquad x^2=\pm4ay

a>0: distance from vertex to focus. y^2=4ax: focus (a,0), directrix x=-a, axis y=0, opens right. x^2=4ay: focus (0,a), directrix y=-a, opens up. A minus sign reverses the opening

Parabola with vertex (h,k)

(y-k)^2=4a(x-h),\qquad (x-h)^2=4a(y-k)

Vertex (h,k); focus (h+a,k) or (h,k+a); directrix x=h-a or y=k-a

Latus rectum of a parabola

\text{length}=|4a|,\qquad \text{ends }(a,\pm2a)\ \text{for }y^2=4ax

The focal chord perpendicular to the axis

Tangent to a parabola

yy_1=2a(x+x_1),\qquad y=mx+\dfrac{a}{m}\ \left(c=\dfrac{a}{m}\right)

For y^2=4ax: tangent at (x_1,y_1), and tangent of slope m with its condition of tangency. Point of contact \left(\dfrac{a}{m^2},\dfrac{2a}{m}\right)

Normal to a parabola at (x_1,y_1)

y-y_1=-\dfrac{y_1}{2a}(x-x_1)

For y^2=4ax; slope of the normal is the negative reciprocal of the tangent slope \dfrac{2a}{y_1}

Ellipse, standard form

\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1\ (a>b),\qquad c^2=a^2-b^2

a: semi-major axis, b: semi-minor axis, c: centre to focus. Foci (\pm c,0), vertices (\pm a,0), co-vertices (0,\pm b). If the larger denominator is under y^2 the major axis is vertical

Ellipse: eccentricity, directrices, latus rectum

e=\dfrac{c}{a}<1,\qquad x=\pm\dfrac{a}{e},\qquad \text{latus rectum}=\dfrac{2b^2}{a}

Sum of focal distances of any point: |PF_1|+|PF_2|=2a

Tangent to an ellipse

\dfrac{xx_1}{a^2}+\dfrac{yy_1}{b^2}=1,\qquad y=mx\pm\sqrt{a^2m^2+b^2}

At the point (x_1,y_1), and in slope form; condition of tangency c^2=a^2m^2+b^2

Normal to an ellipse at (x_1,y_1)

a^2y_1(x-x_1)-b^2x_1(y-y_1)=0

Line through (x_1,y_1) with slope \dfrac{a^2y_1}{b^2x_1}

Hyperbola, standard form

\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1,\qquad c^2=a^2+b^2

2a: transverse axis, 2b: conjugate axis, foci (\pm c,0), vertices (\pm a,0). For \dfrac{y^2}{a^2}-\dfrac{x^2}{b^2}=1 the transverse axis is vertical

Hyperbola: eccentricity, directrices, latus rectum, asymptotes

e=\dfrac{c}{a}>1,\qquad x=\pm\dfrac{a}{e},\qquad \dfrac{2b^2}{a},\qquad y=\pm\dfrac{b}{a}x

Difference of focal distances: \big||PF_1|-|PF_2|\big|=2a. For the vertical hyperbola the asymptotes are y=\pm\dfrac{a}{b}x

Tangent to a hyperbola

\dfrac{xx_1}{a^2}-\dfrac{yy_1}{b^2}=1,\qquad y=mx\pm\sqrt{a^2m^2-b^2}

At (x_1,y_1), and in slope form; condition of tangency c^2=a^2m^2-b^2 (needs |m|>b/a)

Normal to a hyperbola at (x_1,y_1)

a^2y_1(x-x_1)+b^2x_1(y-y_1)=0

Line through (x_1,y_1) with slope -\dfrac{a^2y_1}{b^2x_1}

Shifted conics

x\to x-h,\quad y\to y-k

Centre or vertex moved to (h,k): replace x by x-h and y by y-k in the standard form and add (h,k) to every point (foci, vertices); lines such as directrices shift in the same way

Key terms

Conic section (conic): A curve obtained by cutting a double right circular cone with a plane: circle, ellipse, parabola or hyperbola (plus degenerate cases such as a point or a pair of lines).
Generator: A straight line through the vertex of the cone lying on its surface; the cone is traced by rotating it about the axis.
Nappes: The two parts of a double cone, on either side of the vertex.
Focus: The fixed point used in the locus definition of a conic. An ellipse and a hyperbola have two foci.
Directrix: The fixed line used in the locus definition of a conic; the ratio of a point's distance from the focus to its distance from the directrix is constant.
Eccentricity: The constant ratio e=\dfrac{|PF|}{|PM|}; e=0 for a circle, e<1 ellipse, e=1 parabola, e>1 hyperbola. For an ellipse or hyperbola e=c/a.
Circle: The set of all points in a plane at a fixed distance r (radius) from a fixed point (centre).
Point circle: A circle of radius zero (g^2+f^2-c=0); it is a single point.
Imaginary (virtual) circle: An equation x^2+y^2+2gx+2fy+c=0 with g^2+f^2-c<0; no real point satisfies it.
Secant: A line that cuts a circle (or conic) at two distinct points.
Tangent: A line that meets a curve at exactly one point (the point of contact) without crossing it there; for a circle it is perpendicular to the radius at that point.
Normal: The line through the point of contact perpendicular to the tangent; for a circle it passes through the centre.
Length of tangent: The distance from an external point to the point of contact of a tangent drawn from it, \sqrt{S_1}.
Chord: A line segment joining two points of a curve. A focal chord passes through a focus.
Parabola: The set of all points in a plane equidistant from a fixed point (focus) and a fixed line (directrix) not through the focus; e=1.
Vertex: A point where a conic meets its axis (turning point of a parabola, ends of the major axis of an ellipse or the transverse axis of a hyperbola).
Axis of a parabola: The line through the focus perpendicular to the directrix; the parabola is symmetric about it.
Latus rectum: A focal chord perpendicular to the axis (major or transverse axis). Length 4a for y^2=4ax and \dfrac{2b^2}{a} for an ellipse or hyperbola.
Focal distance: The distance of a point of a conic from a focus.
Ellipse: The set of all points in a plane the sum of whose distances from two fixed points (foci) is a constant 2a; 0<e<1.
Major and minor axes: The longest chord of an ellipse (length 2a, through the foci) and the chord through the centre perpendicular to it (length 2b).
Centre of an ellipse or hyperbola: The midpoint of the line segment joining the foci; the curve is symmetric about it.
Hyperbola: The set of all points in a plane the difference of whose distances from two fixed points (foci) is a constant 2a; e>1.
Transverse and conjugate axes: For a hyperbola, the segment of length 2a joining the vertices (through the foci) and the perpendicular segment of length 2b through the centre.
Asymptote: A line that a branch of the hyperbola approaches more and more closely far from the centre; y=\pm\dfrac{b}{a}x for \dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1.
Condition of tangency: The relation between m and c for which y=mx+c touches a conic, found by setting the discriminant of the combined quadratic to zero.
Reflection property: Rays parallel to the axis of a parabolic mirror reflect through its focus; rays from one focus of an elliptical mirror reflect to the other focus.

Exercise 7.1p. 189–190

Every question asks for a circle. Find the centre and the radius (or g, f, c) from the conditions, then write the standard or general form. Always check the answer with one of the given points.

Find the equation of circle when its centre and radius is given:

Solution

(i) Centre at (3,-1) and radius 2

Use the standard form (x-h)^2+(y-k)^2=r^2 with h=3, k=-1, r=2.
(x-3)^2+(y+1)^2=4
Expand to get the general form.
x^2-6x+9+y^2+2y+1-4=0\;\Longrightarrow\;x^2+y^2-6x+2y+6=0

(x-3)^2+(y+1)^2=4\quad\text{or}\quad x^2+y^2-6x+2y+6=0

(ii) Centre at \left(-\tfrac12,-\tfrac13\right) and radius \tfrac15

Substitute h=-\tfrac12, k=-\tfrac13, r=\tfrac15 in the standard form.
\left(x+\tfrac12\right)^2+\left(y+\tfrac13\right)^2=\tfrac{1}{25}
Expand.
x^2+x+\tfrac14+y^2+\tfrac23y+\tfrac19-\tfrac{1}{25}=0
Multiply by the LCM 900 to clear fractions: 900\left(\tfrac14+\tfrac19-\tfrac1{25}\right)=225+100-36=289.
900x^2+900y^2+900x+600y+289=0

900x^2+900y^2+900x+600y+289=0

(iii) Centre at \left(\tfrac1a,a\right) and radius is a; a\neq0

Substitute h=\tfrac1a, k=a, r=a.
\left(x-\tfrac1a\right)^2+(y-a)^2=a^2
Expand; a^2 cancels.
x^2-\tfrac{2}{a}x+\tfrac{1}{a^2}+y^2-2ay+a^2-a^2=0\;\Longrightarrow\;x^2+y^2-\tfrac2ax-2ay+\tfrac{1}{a^2}=0
Multiply by a^2 to clear the fractions.
a^2x^2+a^2y^2-2ax-2a^3y+1=0

Sketch for a=2: centre \left(\tfrac12,2\right), radius 2. The circle touches the x-axis.

a^2x^2+a^2y^2-2ax-2a^3y+1=0

Convert the following equations of circle into standard form and hence find their centre and radius:

Solution

(i) x^2+y^2-4x+6y-36=0

Group the terms and complete the squares (half of -4 is -2, half of 6 is 3).
\left(x^2-4x+4\right)+\left(y^2+6y+9\right)=36+4+9
Write the squares.
(x-2)^2+(y+3)^2=49=7^2
Compare with (x-h)^2+(y-k)^2=r^2.
\text{Centre }(2,-3),\qquad r=7

(x-2)^2+(y+3)^2=7^2;\ \text{centre }(2,-3),\ r=7

(ii) 5x^2+5y^2-2x+4y-27=0

Divide by 5 so that the coefficients of x^2 and y^2 are 1.
x^2+y^2-\tfrac25x+\tfrac45y-\tfrac{27}{5}=0
Complete the squares: half of -\tfrac25 is -\tfrac15 and half of \tfrac45 is \tfrac25.
\left(x-\tfrac15\right)^2+\left(y+\tfrac25\right)^2=\tfrac{27}{5}+\tfrac{1}{25}+\tfrac{4}{25}=\tfrac{140}{25}=\tfrac{28}{5}
Compare with the standard form.
\text{Centre }\left(\tfrac15,-\tfrac25\right),\qquad r=\sqrt{\tfrac{28}{5}}=\dfrac{2\sqrt{35}}{5}\approx2.37

\left(x-\tfrac15\right)^2+\left(y+\tfrac25\right)^2=\tfrac{28}{5};\ \text{centre }\left(\tfrac15,-\tfrac25\right),\ r=\tfrac{2\sqrt{35}}{5}

(iii) 4x^2+4y^2+2ax+by-a^2=0

Divide by 4.
x^2+y^2+\tfrac{a}{2}x+\tfrac{b}{4}y-\tfrac{a^2}{4}=0
Complete the squares: half of \tfrac{a}{2} is \tfrac{a}{4} and half of \tfrac{b}{4} is \tfrac{b}{8}.
\left(x+\tfrac{a}{4}\right)^2+\left(y+\tfrac{b}{8}\right)^2=\tfrac{a^2}{4}+\tfrac{a^2}{16}+\tfrac{b^2}{64}
Add the fractions over 64.
\left(x+\tfrac{a}{4}\right)^2+\left(y+\tfrac{b}{8}\right)^2=\dfrac{16a^2+4a^2+b^2}{64}=\dfrac{20a^2+b^2}{64}
Compare with the standard form.
\text{Centre }\left(-\tfrac{a}{4},-\tfrac{b}{8}\right),\qquad r=\dfrac{\sqrt{20a^2+b^2}}{8}

Sketch for a=4, b=8: centre (-1,-1), radius \tfrac{\sqrt{384}}{8}\approx2.45.

\text{Centre }\left(-\tfrac{a}{4},-\tfrac{b}{8}\right),\ r=\tfrac{\sqrt{20a^2+b^2}}{8}

32026 Annual I, Set B, Q.3 OR (a)

Find the equation of circle passing through given three non-collinear points.

Solution

(i) (0,2), (2,0), (1,3)

Let the circle be x^2+y^2+2gx+2fy+c=0. Substitute each point.
\begin{aligned}(0,2):&\;\;4+4f+c=0\quad(1)\\(2,0):&\;\;4+4g+c=0\quad(2)\\(1,3):&\;\;10+2g+6f+c=0\quad(3)\end{aligned}
(1) minus (2) gives 4f-4g=0, so f=g. Put f=g in (3) and subtract (2).
(10+8g+c)-(4+4g+c)=0\;\Longrightarrow\;6+4g=0\;\Longrightarrow\;g=-\tfrac32=f
From (2):
c=-4-4g=-4+6=2
Substitute g=f=-\tfrac32, c=2.
x^2+y^2-3x-3y+2=0
Centre and radius.
\left(\tfrac32,\tfrac32\right),\qquad r=\sqrt{\tfrac94+\tfrac94-2}=\dfrac{\sqrt{10}}{2}

x^2+y^2-3x-3y+2=0

(ii) (1,3), (3,6), (5,7)

Substitute the points in x^2+y^2+2gx+2fy+c=0.
\begin{aligned}(1,3):&\;\;2g+6f+c=-10\quad(1)\\(3,6):&\;\;6g+12f+c=-45\quad(2)\\(5,7):&\;\;10g+14f+c=-74\quad(3)\end{aligned}
Subtract to remove c: (2)-(1) and (3)-(2).
4g+6f=-35\quad(4),\qquad 4g+2f=-29\quad(5)
(4)-(5):
4f=-6\;\Longrightarrow\;f=-\tfrac32,\qquad 4g=-29+3\;\Longrightarrow\;g=-\tfrac{13}{2}
From (1):
c=-10-2g-6f=-10+13+9=12
Equation, centre and radius.
x^2+y^2-13x-3y+12=0,\quad\left(\tfrac{13}{2},\tfrac32\right),\quad r=\sqrt{\tfrac{169}{4}+\tfrac94-12}=\dfrac{\sqrt{130}}{2}

x^2+y^2-13x-3y+12=0

(iii) (0,0), (a,0), (0,b); a\neq0, b\neq0

The circle passes through the origin, so c=0: x^2+y^2+2gx+2fy=0.
Put (a,0): a^2+2ga=0. Divide by a\neq0.
g=-\tfrac{a}{2}
Put (0,b): b^2+2fb=0. Divide by b\neq0.
f=-\tfrac{b}{2}
Substitute.
x^2+y^2-ax-by=0
Centre \left(\tfrac{a}{2},\tfrac{b}{2}\right) and radius \tfrac12\sqrt{a^2+b^2}. The angle at the origin is a right angle, so the line joining (a,0) and (0,b) is a diameter.

Sketch for a=4, b=2: the hypotenuse from (a,0) to (0,b) is a diameter.

x^2+y^2-ax-by=0

Find the equation of circle with centre lying on the line x+y=2 and passing through the points (2,-2) and (3,4).

Solution

Let the centre be C(h,k). It lies on x+y=2.
h+k=2\quad(1)
The centre is equally far from A(2,-2) and B(3,4): |CA|^2=|CB|^2.
(h-2)^2+(k+2)^2=(h-3)^2+(k-4)^2
Expand; h^2 and k^2 cancel.
-4h+4+4k+4=-6h+9-8k+16\;\Longrightarrow\;2h+12k=17\quad(2)
From (1), h=2-k. Substitute in (2).
4-2k+12k=17\;\Longrightarrow\;k=\tfrac{13}{10},\qquad h=\tfrac{7}{10}
Radius squared =|CA|^2.
r^2=\left(\tfrac{7}{10}-2\right)^2+\left(\tfrac{13}{10}+2\right)^2=\tfrac{169}{100}+\tfrac{1089}{100}=\tfrac{1258}{100}
Standard form.
\left(x-\tfrac{7}{10}\right)^2+\left(y-\tfrac{13}{10}\right)^2=\tfrac{1258}{100}
Expand: the constant is \tfrac{49+169-1258}{100}=-\tfrac{52}{5}. Multiply by 5.
x^2+y^2-\tfrac75x-\tfrac{13}{5}y-\tfrac{52}{5}=0\;\Longrightarrow\;5x^2+5y^2-7x-13y-52=0
Check: (2,-2) gives 20+20-14+26-52=0 and (3,4) gives 45+80-21-52-52=0.

5x^2+5y^2-7x-13y-52=0

Find the equation of circle passing through the points (1,-2) and (4,-3) and whose centre lies on the line 3x+4y=7.

Solution

Let the centre be C(h,k) with |CA|=|CB| for A(1,-2), B(4,-3).
(h-1)^2+(k+2)^2=(h-4)^2+(k+3)^2
Expand and simplify.
-2h+1+4k+4=-8h+16+6k+9\;\Longrightarrow\;6h-2k=20\;\Longrightarrow\;3h-k=10\quad(1)
The centre lies on the given line.
3h+4k=7\quad(2)
(2)-(1):
5k=-3\;\Longrightarrow\;k=-\tfrac35,\qquad 3h=10-\tfrac35=\tfrac{47}{5}\;\Longrightarrow\;h=\tfrac{47}{15}
Radius squared.
r^2=\left(\tfrac{47}{15}-1\right)^2+\left(-\tfrac35+2\right)^2=\tfrac{1024}{225}+\tfrac{441}{225}=\tfrac{1465}{225}=\tfrac{293}{45}
Standard form.
\left(x-\tfrac{47}{15}\right)^2+\left(y+\tfrac35\right)^2=\tfrac{293}{45}
Expand: the constant is \tfrac{2209+81-1465}{225}=\tfrac{11}{3}. Multiply by 15.
x^2+y^2-\tfrac{94}{15}x+\tfrac65y+\tfrac{11}{3}=0\;\Longrightarrow\;15x^2+15y^2-94x+18y+55=0
Check: (1,-2) gives 15+60-94-36+55=0 and (4,-3) gives 240+135-376-54+55=0.

15x^2+15y^2-94x+18y+55=0

Find the equation of the circle passing through the points (2,1) and touching the line x+2y-1=0 at the point (3,-1).

Solution

P(3,-1) is on the line (3-2-1=0). The radius to the point of contact is perpendicular to the tangent. The tangent has slope -\tfrac12, so the normal at P has slope 2.
y+1=2(x-3)\;\Longrightarrow\;2x-y-7=0
So the centre C(h,k) is on this normal.
2h-k=7\quad(1)
The circle also passes through Q(2,1), so |CP|=|CQ|.
(h-3)^2+(k+1)^2=(h-2)^2+(k-1)^2\;\Longrightarrow\;-6h+2k+10=-4h-2k+5
Simplify.
2h-4k=5\quad(2)
(1)-(2):
3k=2\;\Longrightarrow\;k=\tfrac23,\qquad 2h=7+\tfrac23\;\Longrightarrow\;h=\tfrac{23}{6}
Radius squared =|CP|^2.
r^2=\left(\tfrac{23}{6}-3\right)^2+\left(\tfrac23+1\right)^2=\tfrac{25}{36}+\tfrac{100}{36}=\tfrac{125}{36}
Equation. The constant is \tfrac{529+16-125}{36}=\tfrac{35}{3}; multiply by 3.
\left(x-\tfrac{23}{6}\right)^2+\left(y-\tfrac23\right)^2=\tfrac{125}{36}\;\Longrightarrow\;3x^2+3y^2-23x-4y+35=0
Check the tangency: the distance from C to the line is \dfrac{\left|\tfrac{23}{6}+\tfrac43-1\right|}{\sqrt5}=\dfrac{25/6}{\sqrt5}=\dfrac{5\sqrt5}{6}=r.

3x^2+3y^2-23x-4y+35=0

A circle touches the line 2x-3y+1=0 at point (1,1) and passes through the point of intersection of the lines x+y+1=0 and x-3y+5=0. Find the equation of circle.

Solution

Point of intersection of the two lines: subtract x-3y+5=0 from x+y+1=0.
4y-4=0\;\Longrightarrow\;y=1,\quad x=-2;\qquad Q(-2,1)
The circle touches 2x-3y+1=0 (slope \tfrac23) at P(1,1), so the centre lies on the normal at P, of slope -\tfrac32.
y-1=-\tfrac32(x-1)\;\Longrightarrow\;3x+2y-5=0
So the centre C(h,k) satisfies
3h+2k=5\quad(1)
|CP|=|CQ|:
(h-1)^2+(k-1)^2=(h+2)^2+(k-1)^2\;\Longrightarrow\;-2h+1=4h+4\;\Longrightarrow\;h=-\tfrac12
From (1):
2k=5+\tfrac32\;\Longrightarrow\;k=\tfrac{13}{4}
Radius squared.
r^2=\left(-\tfrac12-1\right)^2+\left(\tfrac{13}{4}-1\right)^2=\tfrac94+\tfrac{81}{16}=\tfrac{117}{16}
Equation. The constant is \tfrac14+\tfrac{169}{16}-\tfrac{117}{16}=\tfrac72; multiply by 2.
\left(x+\tfrac12\right)^2+\left(y-\tfrac{13}{4}\right)^2=\tfrac{117}{16}\;\Longrightarrow\;2x^2+2y^2+2x-13y+7=0
Check: (1,1) gives 2+2+2-13+7=0 and (-2,1) gives 8+2-4-13+7=0.

2x^2+2y^2+2x-13y+7=0

Equations of two diameters of a circle are x-y=3 and 3x+y=5 and its radius is 5. Find the equation of circle.

Solution

Every diameter passes through the centre, so the centre is the point of intersection of the two diameters. Add the equations.
4x=8\;\Longrightarrow\;x=2,\qquad y=x-3=-1
Centre (2,-1) and r=5.
(x-2)^2+(y+1)^2=25
Expand.
x^2+y^2-4x+2y-20=0

x^2+y^2-4x+2y-20=0

A circle touches both the axes in the first quadrant and area of the circle is 13\pi square units. Find the equation of the circle.

Solution

Area \pi r^2=13\pi.
r^2=13\;\Longrightarrow\;r=\sqrt{13}
A circle touching both axes has its centre at distance r from each axis. In the first quadrant the centre is
(r,r)=\left(\sqrt{13},\sqrt{13}\right)
Standard form and expansion.
\left(x-\sqrt{13}\right)^2+\left(y-\sqrt{13}\right)^2=13\;\Longrightarrow\;x^2+y^2-2\sqrt{13}\,x-2\sqrt{13}\,y+13=0

x^2+y^2-2\sqrt{13}\,x-2\sqrt{13}\,y+13=0

The two points A(4,3) and B(2,5) lie on the circle and the centre of the circle lies on the perpendicular bisector of the chord \overline{AB}. The distance between the centre and the chord \overline{AB} is \sqrt7. Find the equation of the circle.

Solution

Midpoint of \overline{AB} and half its length.
M=\left(\tfrac{4+2}{2},\tfrac{3+5}{2}\right)=(3,4),\qquad|AM|=\sqrt{(4-3)^2+(3-4)^2}=\sqrt2
The perpendicular from the centre C to the chord meets it at M, so triangle CMA is right-angled at M.
r^2=|CM|^2+|AM|^2=7+2=9\;\Longrightarrow\;r=3
Slope of AB is \tfrac{5-3}{2-4}=-1, so the perpendicular bisector has slope 1 and passes through M.
y-4=x-3\;\Longrightarrow\;y=x+1
Let C(h,h+1). Its distance from M(3,4) is \sqrt7.
(h-3)^2+(h-3)^2=7\;\Longrightarrow\;h-3=\pm\sqrt{\tfrac72}=\pm\dfrac{\sqrt{14}}{2}
So there are two possible centres, one on each side of the chord.
C_1\left(3+\tfrac{\sqrt{14}}{2},\,4+\tfrac{\sqrt{14}}{2}\right),\qquad C_2\left(3-\tfrac{\sqrt{14}}{2},\,4-\tfrac{\sqrt{14}}{2}\right)
Equations with r^2=9.
\left(x-3\mp\tfrac{\sqrt{14}}{2}\right)^2+\left(y-4\mp\tfrac{\sqrt{14}}{2}\right)^2=9
Expanded: the constant is h^2+k^2-9=23\pm7\sqrt{14}.
x^2+y^2-\left(6\pm\sqrt{14}\right)x-\left(8\pm\sqrt{14}\right)y+23\pm7\sqrt{14}=0
Check: the distance of C_1 from the chord x+y-7=0 is \dfrac{\left|7+\sqrt{14}-7\right|}{\sqrt2}=\sqrt7.

x^2+y^2-\left(6\pm\sqrt{14}\right)x-\left(8\pm\sqrt{14}\right)y+23\pm7\sqrt{14}=0

Find the equation of the circle passing through the intersection of the circles C_1:\ x^2+y^2-8x-2y+7=0 and C_2:\ x^2+y^2-4x+10y+8=0 and passes through (-1,-2). (Hint: equation of the required circle is C_1+\lambda C_2=0; \lambda\neq-1)

Solution

Any circle through the points common to C_1=0 and C_2=0 has the form C_1+\lambda C_2=0, because both points make C_1 and C_2 zero. (\lambda=-1 is excluded: it removes x^2 and y^2 and gives a line.)
\begin{aligned}&\left(x^2+y^2-8x-2y+7\right)\\ &\quad+\lambda\left(x^2+y^2-4x+10y+8\right)=0\end{aligned}
The circle passes through (-1,-2). Evaluate each bracket there.
\begin{aligned}C_1(-1,-2)&=1+4+8+4+7=24\\ C_2(-1,-2)&=1+4+4-20+8=-3\end{aligned}
So
24+\lambda(-3)=0\;\Longrightarrow\;\lambda=8
Substitute \lambda=8 and collect terms.
9x^2+9y^2-40x+78y+71=0
Check: (-1,-2) gives 9+36+40-156+71=0. The two given circles do meet: their centres (4,1) and (2,-5) are \sqrt{40}\approx6.3 apart, which lies between \left|\sqrt{21}-\sqrt{10}\right|\approx1.4 and \sqrt{21}+\sqrt{10}\approx7.7.

The required circle (solid) passes through both points where C_1 and C_2 meet, and through (-1,-2).

9x^2+9y^2-40x+78y+71=0

The diagram shows a rectangle PQRS and the circles C_1 and C_2. Both the circles touch each other and three sides of the rectangle. The coordinates of the points P, Q, R and S are (0,4), (1,1), (7,3) and (6,6). Find the equation of the circles C_1 and C_2.

The rectangle PQRS and the circles, drawn from the given coordinates. (The book's sketch places the letters differently, but the coordinates decide the answer.)

Solution

Side lengths. \overrightarrow{PQ}=(1,-3) and \overrightarrow{QR}=(6,2); their dot product is 6-6=0, so the corner at Q is a right angle.
|PQ|=|RS|=\sqrt{10},\qquad|QR|=|SP|=\sqrt{40}=2\sqrt{10}
Each circle touches both long sides QR and SP, so its diameter equals the distance between them, |PQ|=\sqrt{10}.
r=\dfrac{\sqrt{10}}{2},\qquad r^2=\dfrac{10}{4}=\dfrac52
The centres lie on the midline joining the midpoints of PQ and RS.
M_1=\left(\tfrac12,\tfrac52\right),\qquad M_2=\left(\tfrac{13}{2},\tfrac92\right)
C_1 touches PQ, so its centre is a distance r from M_1 along the midline. The unit vector along the midline is \dfrac{(6,2)}{2\sqrt{10}}=\dfrac{(3,1)}{\sqrt{10}}.
\text{centre of }C_1=\left(\tfrac12,\tfrac52\right)+\dfrac{\sqrt{10}}{2}\cdot\dfrac{(3,1)}{\sqrt{10}}=(2,3)
C_2 touches RS, so its centre is a distance r from M_2 back along the midline.
\text{centre of }C_2=\left(\tfrac{13}{2},\tfrac92\right)-\left(\tfrac32,\tfrac12\right)=(5,4)
Check that they touch each other: the distance between the centres equals 2r.
\sqrt{(5-2)^2+(4-3)^2}=\sqrt{10}=2r
Check one side: PQ is 3x+y-4=0 and the distance of (2,3) from it is \dfrac{|6+3-4|}{\sqrt{10}}=\dfrac{5}{\sqrt{10}}=\dfrac{\sqrt{10}}{2}=r.
Equations (multiplied by 2).
C_1:\ (x-2)^2+(y-3)^2=\tfrac52\;\Longrightarrow\;2x^2+2y^2-8x-12y+21=0
C_2:\ (x-5)^2+(y-4)^2=\tfrac52\;\Longrightarrow\;2x^2+2y^2-20x-16y+77=0

\begin{aligned}C_1&:\ 2x^2+2y^2-8x-12y+21=0\\ C_2&:\ 2x^2+2y^2-20x-16y+77=0\end{aligned}

Note. With the given coordinates, PQ and RS are the short sides. The circle next to PQ (lower left in the coordinate plane) is taken as C_1 and the other as C_2, as in the book's diagram.

A circle has its centre at the point C(0,1) and a line touches the circle. The point P(3,5) lies on the line touching the circle. The distance between P and C is five times the radius of the circle. Find the equation of the circle and the point where the line touches the circle.

Solution

Distance |PC|.
|PC|=\sqrt{(3-0)^2+(5-1)^2}=\sqrt{25}=5
|PC|=5r, so
r=1
Equation of the circle with centre (0,1) and radius 1.
x^2+(y-1)^2=1\;\Longrightarrow\;x^2+y^2-2y=0\quad(1)
Let the point of contact be T(x,y). The radius CT is perpendicular to the tangent PT, so \overrightarrow{CT}\cdot\overrightarrow{PT}=0.
x(x-3)+(y-1)(y-5)=0\;\Longrightarrow\;x^2+y^2-3x-6y+5=0\quad(2)
Subtract (1) from (2) to get a straight line through the points of contact.
-3x-4y+5=0\;\Longrightarrow\;x=\dfrac{5-4y}{3}
Substitute in (1) and multiply by 9.
(5-4y)^2+9y^2-18y=0\;\Longrightarrow\;25y^2-58y+25=0
Solve the quadratic: \sqrt{58^2-4(25)(25)}=\sqrt{864}=12\sqrt6.
y=\dfrac{58\pm12\sqrt6}{50}=\dfrac{29\pm6\sqrt6}{25},\qquad x=\dfrac{5-4y}{3}=\dfrac{3\mp8\sqrt6}{25}
Two tangents can be drawn from P, so there are two possible points of contact.
T_1\left(\tfrac{3-8\sqrt6}{25},\tfrac{29+6\sqrt6}{25}\right)\approx(-0.66,\,1.75),\qquad T_2\left(\tfrac{3+8\sqrt6}{25},\tfrac{29-6\sqrt6}{25}\right)\approx(0.90,\,0.57)
Check: the tangent length is \sqrt{|PC|^2-r^2}=\sqrt{24}, and |PT_2|^2=(3-0.904)^2+(5-0.572)^2\approx24.0.

x^2+y^2-2y=0;\quad T\left(\tfrac{3\mp8\sqrt6}{25},\,\tfrac{29\pm6\sqrt6}{25}\right)

Note. The question speaks of one tangent line, but two tangents can be drawn from P, so both points of contact are given. The tangents have slopes \tfrac{6\pm\sqrt6}{4}.

The three lines 2x-y+1=0; 2x+y-3=0 and x-2y+4=0 touch the circle. Find the centre of the circle. Also find the equation of circle.

Solution

The centre C(h,k) is at the same distance r from all three lines. Each line has \sqrt{a^2+b^2}=\sqrt5, so the numerators must be equal in size.
|2h-k+1|=|2h+k-3|=|h-2k+4|=\sqrt5\,r
The lines form a small triangle with vertices
\left(\tfrac12,2\right),\quad\left(\tfrac23,\tfrac73\right),\quad\left(\tfrac25,\tfrac{11}{5}\right)
For the inscribed circle (inside the triangle) take the signs that hold at an inside point, for example the centroid \left(0.52,\,2.18\right): there 2h-k+1<0, 2h+k-3>0 and h-2k+4>0.
-(2h-k+1)=2h+k-3=h-2k+4
The first equality gives 4h=2; then the second gives k.
h=\tfrac12,\qquad 1+k-3=\tfrac12-2k+4\;\Longrightarrow\;3k=\tfrac{13}{2}\;\Longrightarrow\;k=\tfrac{13}{6}
Radius.
r=\dfrac{\left|2\left(\tfrac12\right)+\tfrac{13}{6}-3\right|}{\sqrt5}=\dfrac{1/6}{\sqrt5}=\dfrac{1}{6\sqrt5},\qquad r^2=\dfrac{1}{180}
Equation of the circle; multiply by 180.
\left(x-\tfrac12\right)^2+\left(y-\tfrac{13}{6}\right)^2=\tfrac{1}{180}\;\Longrightarrow\;180x^2+180y^2-180x-780y+889=0
The other sign choices give the three escribed circles, which touch one side and the other two sides extended.
\begin{aligned}&(1,2),\ r^2=\tfrac15:\\ &\quad 5x^2+5y^2-10x-20y+24=0\\[4pt] &\left(\tfrac13,2\right),\ r^2=\tfrac{1}{45}:\\ &\quad 45x^2+45y^2-30x-180y+184=0\\[4pt] &\left(\tfrac12,\tfrac52\right),\ r^2=\tfrac{1}{20}:\\ &\quad 20x^2+20y^2-20x-100y+129=0\end{aligned}

The three lines make a very small triangle. The incircle (solid) has centre \left(\tfrac12,\tfrac{13}{6}\right); the three escribed circles are dashed.

\text{Centre }\left(\tfrac12,\tfrac{13}{6}\right),\quad 180x^2+180y^2-180x-780y+889=0

Note. Four circles touch three lines that form a triangle. The inscribed circle is the usual answer; the three escribed circles are also listed.

Exercise 7.2p. 192

Line and circle problems: substitute the line into the circle and look at the discriminant, or compare the distance of the centre from the line with the radius. For tangents and normals at a point, first check that the point lies on the circle.

Find the points of intersection of the given line and the circle.

Solution

(i) x-y+1=0; x^2+y^2-3x-8=0

Make y the subject of the linear equation, so it can be substituted in the circle.
y=x+1
Substitute in the circle and simplify.
x^2+(x+1)^2-3x-8=0\;\Longrightarrow\;2x^2-x-7=0
Solve by the quadratic formula. The discriminant 1+56=57>0, so there are two points.
x=\dfrac{1\pm\sqrt{57}}{4}
Find y from y=x+1.
y=\dfrac{1\pm\sqrt{57}}{4}+1=\dfrac{5\pm\sqrt{57}}{4}
So the line cuts the circle at two points, approximately (2.14,\,3.14) and (-1.64,\,-0.64).

\left(\dfrac{1+\sqrt{57}}{4},\dfrac{5+\sqrt{57}}{4}\right),\ \left(\dfrac{1-\sqrt{57}}{4},\dfrac{5-\sqrt{57}}{4}\right)

(ii) 2x+y+4\sqrt5=0; x^2+y^2-2x+4y-11=0

From the line:
y=-2x-4\sqrt5
Substitute in the circle. Note (-2x-4\sqrt5)^2=4x^2+16\sqrt5\,x+80.
x^2+4x^2+16\sqrt5\,x+80-2x-8x-16\sqrt5-11=0
Collect terms.
5x^2+\left(16\sqrt5-10\right)x+\left(69-16\sqrt5\right)=0
Discriminant: \left(16\sqrt5-10\right)^2-20\left(69-16\sqrt5\right)=1380-320\sqrt5-1380+320\sqrt5=0. Equal roots, so the line touches the circle.
x=\dfrac{10-16\sqrt5}{10}=1-\dfrac{8\sqrt5}{5}
Find y.
y=-2\left(1-\dfrac{8\sqrt5}{5}\right)-4\sqrt5=-2+\dfrac{16\sqrt5}{5}-\dfrac{20\sqrt5}{5}=-2-\dfrac{4\sqrt5}{5}
Check with distances: centre (1,-2), r=\sqrt{1+4+11}=4, and the distance of the centre from the line is \dfrac{|2-2+4\sqrt5|}{\sqrt5}=4=r.

The line is a tangent: it meets the circle at one point only.

\left(1-\dfrac{8\sqrt5}{5},\ -2-\dfrac{4\sqrt5}{5}\right)\approx(-2.58,\,-3.79)\ \text{(one point: tangent)}

(iii) 3x+2y+1=0; x^2+y^2-x+y+2=0

From the line:
y=\dfrac{-1-3x}{2}
Substitute and multiply by 4.
4x^2+(1+3x)^2-4x-2(1+3x)+8=0\;\Longrightarrow\;13x^2-4x+7=0
Discriminant.
b^2-4ac=16-4(13)(7)=-348<0
No real roots, so there are no points of intersection.
Reason: here g=-\tfrac12, f=\tfrac12, c=2, so g^2+f^2-c=\tfrac14+\tfrac14-2=-\tfrac32<0. The equation is an imaginary (virtual) circle: no real point satisfies it, so no line can meet it.

\text{No point of intersection}

Note. In part (iii) the given equation does not represent a real circle (g^2+f^2-c<0), which is why there is no intersection.

The equation of a circle is x^2+y^2-4x+y-7=0 and the equation of line is 2x-y+c=0. Find the value(s) of c such that the line: (i) intersects the circle at two distinct points. (ii) is tangent to circle. (iii) has no common point with the circle.

Solution

Centre and radius: g=-2, f=\tfrac12, c_0=-7.
C\left(2,-\tfrac12\right),\qquad r=\sqrt{4+\tfrac14+7}=\sqrt{\tfrac{45}{4}}=\dfrac{3\sqrt5}{2}
Distance of the centre from the line 2x-y+c=0.
d=\dfrac{\left|2(2)-\left(-\tfrac12\right)+c\right|}{\sqrt{4+1}}=\dfrac{\left|c+\tfrac92\right|}{\sqrt5}
Boundary case d=r (tangent).
\dfrac{\left|c+\tfrac92\right|}{\sqrt5}=\dfrac{3\sqrt5}{2}\;\Longrightarrow\;\left|c+\tfrac92\right|=\tfrac{15}{2}\;\Longrightarrow\;c=3\ \text{or}\ c=-12
Same result by substitution: put y=2x+c in the circle.
\begin{aligned}&5x^2+(4c-2)x+\left(c^2+c-7\right)=0\\ \Delta&=(4c-2)^2-20\left(c^2+c-7\right)\\ &=-4(c-3)(c+12)\end{aligned}
\Delta>0 (two points) when (c-3)(c+12)<0, that is between the roots; \Delta<0 outside the roots.

Answers

(i) Two distinct points: d<r, that is -12<c<3.
(ii) Tangent: d=r, that is c=3 or c=-12.
(iii) No common point: d>r, that is c<-12 or c>3.

All lines y=2x+c are parallel. Those between the two tangents (c=3 and c=-12) cut the circle; those outside miss it.

\text{(i) }-12<c<3,\quad\text{(ii) }c=3\text{ or }-12,\quad\text{(iii) }c<-12\text{ or }c>3

The tangents to a circle (x-1)^2+(y-2)^2=3^2 are perpendicular to the line x+3y-6=0. Find the equations of the tangent lines.

Solution

The circle has centre (1,2) and r=3. The given line has slope -\tfrac13, so a perpendicular line has slope 3.
m=3
Every line of slope 3 can be written with one unknown constant k.
3x-y+k=0
For a tangent, the distance from the centre equals the radius.
\dfrac{|3(1)-2+k|}{\sqrt{9+1}}=3\;\Longrightarrow\;|1+k|=3\sqrt{10}
So
k=-1\pm3\sqrt{10}
The two tangents:
3x-y-1+3\sqrt{10}=0\quad\text{and}\quad3x-y-1-3\sqrt{10}=0

y=3x-1\pm3\sqrt{10}\quad\text{i.e. }3x-y-1\pm3\sqrt{10}=0

Find the equations of tangent and normal to the circle x^2+y^2+3x+2y+3=0 at the point (2,-1).

Solution

Checking the point

A tangent at a point needs the point to lie on the circle. Substitute (2,-1).
4+1+6-2+3=12\neq0
The value is positive, so (2,-1) lies outside this circle (centre \left(-\tfrac32,-1\right), r=\tfrac12). There is a misprint. Two small corrections make the question work, and both are solved below.

Version 1: circle x^2+y^2-3x+2y+3=0 (sign of 3x corrected)

Check: 4+1-6-2+3=0, so (2,-1) is on this circle. Here g=-\tfrac32, f=1, c=3; centre \left(\tfrac32,-1\right), r=\sqrt{\tfrac94+1-3}=\tfrac12.
Tangent at (x_1,y_1): xx_1+yy_1+g(x+x_1)+f(y+y_1)+c=0.
2x-y-\tfrac32(x+2)+(y-1)+3=0\;\Longrightarrow\;\tfrac12x-1=0\;\Longrightarrow\;x=2
The normal passes through the point and the centre \left(\tfrac32,-1\right). Both have y=-1.
y=-1\quad\text{i.e. }y+1=0

Version 2: point (-2,-1) (sign of the x-coordinate corrected)

Check on the printed circle: 4+1-6-2+3=0, so (-2,-1) is on it. Centre \left(-\tfrac32,-1\right), g=\tfrac32, f=1, c=3.
Tangent formula with (x_1,y_1)=(-2,-1).
-2x-y+\tfrac32(x-2)+(y-1)+3=0\;\Longrightarrow\;-\tfrac12x-1=0\;\Longrightarrow\;x=-2
Normal through (-2,-1) and the centre \left(-\tfrac32,-1\right).
y=-1

Both versions: the point is an end of a horizontal diameter, so the tangent is vertical and the normal is the line y=-1.

\text{Tangent }x=2,\ \text{normal }y=-1\ \ (\text{or }x=-2,\ y=-1\text{ for the point }(-2,-1))

Note. The point (2,-1) does not lie on the printed circle x^2+y^2+3x+2y+3=0. Either the circle should be x^2+y^2-3x+2y+3=0 or the point should be (-2,-1); the answers are x-2=0,\ y+1=0 and x+2=0,\ y+1=0 respectively.

Find the equation of the tangent to the circle 36x^2+36y^2-72y+11=0 at the point on the circle with abscissa \tfrac12.

Solution

Find the points first: put x=\tfrac12 (abscissa means x-coordinate).
9+36y^2-72y+11=0\;\Longrightarrow\;36y^2-72y+20=0\;\Longrightarrow\;9y^2-18y+5=0
Factorise.
(3y-5)(3y-1)=0\;\Longrightarrow\;y=\tfrac53\ \text{or}\ y=\tfrac13
Divide the circle by 36: x^2+y^2-2y+\tfrac{11}{36}=0, so g=0, f=-1, c=\tfrac{11}{36}; centre (0,1), r=\sqrt{1-\tfrac{11}{36}}=\tfrac56.
Tangent at \left(\tfrac12,\tfrac53\right): use xx_1+yy_1+f(y+y_1)+c=0 and multiply by 36.
\tfrac12x+\tfrac53y-\left(y+\tfrac53\right)+\tfrac{11}{36}=0\;\Longrightarrow\;18x+24y-49=0
Tangent at \left(\tfrac12,\tfrac13\right) in the same way.
\tfrac12x+\tfrac13y-\left(y+\tfrac13\right)+\tfrac{11}{36}=0\;\Longrightarrow\;18x-24y-1=0
Check: 18\left(\tfrac12\right)+24\left(\tfrac53\right)-49=9+40-49=0 and 9-8-1=0.

18x+24y-49=0\quad\text{and}\quad18x-24y-1=0

Find the equation of the normal to the circle x^2+y^2-2x+2y-11=0 at the point on the circle with ordinate -4.

Solution

Put y=-4 (ordinate means y-coordinate) to find the points.
x^2+16-2x-8-11=0\;\Longrightarrow\;x^2-2x-3=0\;\Longrightarrow\;(x-3)(x+1)=0
So the points are (3,-4) and (-1,-4). The centre is (-g,-f)=(1,-1) and every normal passes through the centre.
Normal at (3,-4): slope of the line joining it to (1,-1) is \dfrac{-4+1}{3-1}=-\tfrac32.
y+1=-\tfrac32(x-1)\;\Longrightarrow\;3x+2y-1=0
Normal at (-1,-4): slope \dfrac{-4+1}{-1-1}=\tfrac32.
y+1=\tfrac32(x-1)\;\Longrightarrow\;3x-2y-5=0

3x+2y-1=0\quad\text{and}\quad3x-2y-5=0

Circles with equations x^2+y^2-2y-3=0 and x^2+y^2-8x+4y+11=0 touch each other externally. Find the equation of the common tangent to the given circles.

Solution

Centres and radii.
\begin{aligned}C_1:&\ (0,1),\ r_1=\sqrt{0+1+3}=2\\C_2:&\ (4,-2),\ r_2=\sqrt{16+4-11}=3\end{aligned}
They touch externally because the distance between the centres equals the sum of the radii.
|C_1C_2|=\sqrt{16+9}=5=r_1+r_2
At the point of contact both circle equations are satisfied. Subtracting them removes x^2+y^2 and leaves a straight line through the common point: this is the common tangent there.
(x^2+y^2-2y-3)-(x^2+y^2-8x+4y+11)=0\;\Longrightarrow\;8x-6y-14=0
Simplify.
4x-3y-7=0
Point of contact: it divides C_1C_2 in the ratio r_1:r_2=2:3.
P=\left(\dfrac{2(4)+3(0)}{5},\dfrac{2(-2)+3(1)}{5}\right)=\left(\tfrac85,-\tfrac15\right)
Check: the tangent must be perpendicular to C_1C_2. Slope of C_1C_2=-\tfrac34 and slope of 4x-3y-7=0 is \tfrac43; the product is -1. Also P satisfies 4\left(\tfrac85\right)-3\left(-\tfrac15\right)-7=0.

4x-3y-7=0

Note. The book asks for the tangent at the point of contact. Two circles that touch externally also have two direct common tangents that do not pass between them: y=mx+c with m=\dfrac{-12\mp2\sqrt6}{15}, c=\dfrac{9\mp16\sqrt6}{15} (about y=-1.13x-2.01 and y=-0.47x+3.21).

Show that the tangent line at any point P on the circle is always perpendicular to the radial line through point P.

Solution

Take the general circle x^2+y^2+2gx+2fy+c=0 with centre C(-g,-f) and a point P(x_1,y_1) on it. We find the slope of the tangent at P by differentiation and compare it with the slope of CP.

Differentiate the circle implicitly with respect to x.
2x+2y\dfrac{dy}{dx}+2g+2f\dfrac{dy}{dx}=0\;\Longrightarrow\;\dfrac{dy}{dx}=-\dfrac{x+g}{y+f}
Slope of the tangent at P(x_1,y_1).
m_1=-\dfrac{x_1+g}{y_1+f}
Slope of the radial line CP, joining C(-g,-f) to P(x_1,y_1).
m_2=\dfrac{y_1-(-f)}{x_1-(-g)}=\dfrac{y_1+f}{x_1+g}
Multiply the slopes.
m_1m_2=-\dfrac{x_1+g}{y_1+f}\cdot\dfrac{y_1+f}{x_1+g}=-1
The product of the slopes is -1, so the tangent at P is perpendicular to CP.
Special cases: if y_1+f=0, P is level with the centre, CP is horizontal and the tangent is vertical; if x_1+g=0, CP is vertical and the tangent is horizontal. They are perpendicular in these cases too.

The radius CP and the tangent at P meet at a right angle.

m_{\text{tangent}}\cdot m_{CP}=-1\ \Rightarrow\ \text{tangent}\perp CP

A(-5,-1) and B(1,5) are two points on the circle x^2+y^2=26. Find the point of intersection of the tangents to the circle at A and B. Show that the point of intersection of the tangent lines, the midpoint of chord \overline{AB} and the centre of the circle are collinear.

Solution

Check the points: 25+1=26 and 1+25=26, so both lie on the circle.
The tangent to x^2+y^2=r^2 at (x_1,y_1) is xx_1+yy_1=r^2.
\text{at }A:\ -5x-y=26\quad(1),\qquad\text{at }B:\ x+5y=26\quad(2)
From (2), x=26-5y. Substitute in (1).
-130+25y-y=26\;\Longrightarrow\;24y=156\;\Longrightarrow\;y=\tfrac{13}{2},\quad x=26-\tfrac{65}{2}=-\tfrac{13}{2}
So the tangents meet at
T\left(-\tfrac{13}{2},\tfrac{13}{2}\right)
Midpoint of the chord \overline{AB}; the centre is O(0,0).
M=\left(\dfrac{-5+1}{2},\dfrac{-1+5}{2}\right)=(-2,2)
Collinearity: compare slopes of OM and OT.
m_{OM}=\dfrac{2-0}{-2-0}=-1,\qquad m_{OT}=\dfrac{13/2}{-13/2}=-1
Equal slopes through the common point O, so O, M and T lie on the line y=-x. Equivalently the area determinant is zero.
\begin{vmatrix}0&0&1\\-2&2&1\\-\tfrac{13}{2}&\tfrac{13}{2}&1\end{vmatrix}=0\cdot(\ldots)-0+\left(-13+13\right)=0

T\left(-\tfrac{13}{2},\tfrac{13}{2}\right);\ O,\ M(-2,2),\ T\ \text{lie on } y=-x

A normal line cuts the circle with centre at (1,2) at the point (3,5). Find the other point of intersection of the normal and the circle. Also find the equation of the circle.

Solution

Every normal of a circle passes through its centre, so the normal through (3,5) is a diameter. The other end of the diameter is the required point Q(x,y), and the centre (1,2) is the midpoint of the diameter.
\dfrac{3+x}{2}=1,\qquad\dfrac{5+y}{2}=2
Solve.
x=-1,\qquad y=-1\;\Longrightarrow\;Q(-1,-1)
Radius = distance from the centre to (3,5).
r^2=(3-1)^2+(5-2)^2=13
Equation of the circle.
(x-1)^2+(y-2)^2=13\;\Longrightarrow\;x^2+y^2-2x-4y-8=0
Check Q: 1+1+2+4-8=0. The normal itself is the line through (3,5) and (1,2): slope \tfrac32, so 3x-2y+1=0.

(-1,-1);\quad x^2+y^2-2x-4y-8=0

Exercise 7.3p. 194

Position of a point with respect to a circle (Section 7.5) and the length of a tangent from an outside point (Section 7.6). In both, first make the coefficients of x^2 and y^2 equal to 1.

Check whether the given point lies inside, outside or on the given circle

Solution

Rule (Section 7.5). First make the coefficients of x^2 and y^2 equal to 1, so the circle is x^2+y^2+2gx+2fy+c=0. Then put the point P(x_1,y_1) into the left-hand side:

x_1^2+y_1^2+2gx_1+2fy_1+c\;\begin{cases}>0 & \Rightarrow\ P\text{ lies outside the circle}\\ =0 & \Rightarrow\ P\text{ lies on the circle}\\ <0 & \Rightarrow\ P\text{ lies inside the circle}\end{cases}

(i) (3,5); x^2+y^2-6x+2y-18=0

The coefficients of x^2 and y^2 are already 1, so we can substitute the point directly into the left-hand side.
Put x=3, y=5.
(3)^2+(5)^2-6(3)+2(5)-18=9+25-18+10-18=8
The value is positive, so the point is outside.
8>0\;\Longrightarrow\;(3,5)\ \text{lies outside the circle}
Check with distances. Here g=-3, f=1, c=-18, so the centre is C(3,-1) and r=\sqrt{9+1+18}=\sqrt{28}\approx5.29. The distance from the centre is |CP|=\sqrt{0^2+6^2}=6>r, which agrees.

P is 6 units from the centre but the radius is only about 5.29, so P is outside.

8>0:\ (3,5)\ \text{lies outside the circle}

(ii) (1,6); 2x^2+2y^2-4x-16y+5=0

Make the coefficients of x^2 and y^2 equal to 1 by dividing the equation by 2.
x^2+y^2-2x-8y+\dfrac52=0
Put x=1, y=6.
(1)^2+(6)^2-2(1)-8(6)+\dfrac52=1+36-2-48+\dfrac52=-13+\dfrac52=-\dfrac{21}{2}
The value is negative, so the point is inside.
-\dfrac{21}{2}<0\;\Longrightarrow\;(1,6)\ \text{lies inside the circle}
Check with distances. Centre C(1,4) and r^2=1+16-\tfrac52=\tfrac{29}{2}, so r\approx3.81. Then |CP|=\sqrt{0^2+2^2}=2<r, which agrees.

-\tfrac{21}{2}<0:\ (1,6)\ \text{lies inside the circle}

(iii) (-1,-2); x^2+y^2+6x+4y+9=0

The coefficients of x^2 and y^2 are 1. Put x=-1, y=-2 into the left-hand side.
(-1)^2+(-2)^2+6(-1)+4(-2)+9=1+4-6-8+9=0
The value is zero, so the point satisfies the equation of the circle.
0=0\;\Longrightarrow\;(-1,-2)\ \text{lies on the circle}
Check with distances. Centre C(-3,-2) and r=\sqrt{9+4-9}=2. The distance |CP|=\sqrt{2^2+0^2}=2=r, which agrees.

0:\ (-1,-2)\ \text{lies on the circle}

Find the length of the tangent to the circle x^2+y^2-18x+16y+10=0 from the point (-1,-1) lying outside the circle.

Solution

Formula (Section 7.6). For the circle x^2+y^2+2gx+2fy+c=0 (coefficients of x^2 and y^2 equal to 1), the length of the tangent from P(x_1,y_1) is

|AP|=\sqrt{x_1^2+y_1^2+2gx_1+2fy_1+c}

The coefficients of x^2 and y^2 are 1. Compare with the general form.
2g=-18,\quad 2f=16,\quad c=10,\qquad (x_1,y_1)=(-1,-1)
Substitute the point into the left-hand side of the circle.
x_1^2+y_1^2-18x_1+16y_1+10=(-1)^2+(-1)^2-18(-1)+16(-1)+10
Simplify.
=1+1+18-16+10=14
Take the square root.
|AP|=\sqrt{14}\approx3.74\ \text{units}
Check with Pythagoras. Centre C(9,-8), r^2=81+64-10=135 and |CP|^2=(9+1)^2+(-8+1)^2=149. The radius CA is perpendicular to the tangent, so |AP|^2=|CP|^2-r^2=149-135=14, which agrees.

Part of the circle near P (radius \sqrt{135}\approx11.6). The two tangents from P have the same length \sqrt{14}.

\text{Length of tangent}=\sqrt{14}\approx3.74\ \text{units}

Find the length of tangent to the circle 3x^2+3y^2+18x-24y+50=0 drawn from a point (-3,1) lying outside the circle.

Solution

The formula \sqrt{x_1^2+y_1^2+2gx_1+2fy_1+c} needs the coefficients of x^2 and y^2 to be 1, so divide the equation by 3.
x^2+y^2+6x-8y+\dfrac{50}{3}=0
Substitute (x_1,y_1)=(-3,1) into the left-hand side.
(-3)^2+(1)^2+6(-3)-8(1)+\dfrac{50}{3}=9+1-18-8+\dfrac{50}{3}
Simplify.
=-16+\dfrac{50}{3}=\dfrac{-48+50}{3}=\dfrac{2}{3}
Take the square root and rationalise.
|AP|=\sqrt{\dfrac23}=\dfrac{\sqrt6}{3}\approx0.82\ \text{units}
Check. Centre C(-3,4), r^2=9+16-\tfrac{50}{3}=\tfrac{25}{3} and |CP|^2=0^2+3^2=9. So |AP|^2=9-\tfrac{25}{3}=\tfrac23, which agrees.

P is just outside the circle, so the tangents are short: |AP|=\sqrt6/3.

\text{Length of tangent}=\sqrt{\tfrac23}=\dfrac{\sqrt6}{3}\approx0.82\ \text{units}

The point P(-11,-10) lies outside the circle x^2+y^2+6x+8y+5=0. Find the equations of the tangents to the circle drawn from point P. Also find the points of contact.

Solution

Centre and radius. Compare with x^2+y^2+2gx+2fy+c=0: g=3, f=4, c=5.
\begin{aligned}C(-g,-f)&=(-3,-4)\\ r&=\sqrt{g^2+f^2-c}=\sqrt{9+16-5}=\sqrt{20}=2\sqrt5\end{aligned}
**P is outside** (as stated): substituting P gives a positive value.
(-11)^2+(-10)^2+6(-11)+8(-10)+5=121+100-66-80+5=80>0
**A line through P.** Let the slope be m and use the point-slope form.
y+10=m(x+11)\;\Longrightarrow\;mx-y+(11m-10)=0
Condition of tangency. A line touches a circle when its perpendicular distance from the centre equals the radius (Section 7.3). Use d=\dfrac{|ax_0+by_0+c|}{\sqrt{a^2+b^2}} with the centre (-3,-4).
\dfrac{|m(-3)-(-4)+11m-10|}{\sqrt{m^2+1}}=2\sqrt5\;\Longrightarrow\;\dfrac{|8m-6|}{\sqrt{m^2+1}}=2\sqrt5
Square both sides to remove the modulus and the root.
(8m-6)^2=20(m^2+1)\;\Longrightarrow\;64m^2-96m+36=20m^2+20\;\Longrightarrow\;44m^2-96m+16=0
Divide by 4 and factorise.
11m^2-24m+4=0\;\Longrightarrow\;(m-2)(11m-2)=0\;\Longrightarrow\;m=2\ \text{ or }\ m=\dfrac{2}{11}
The vertical line x=-11 through P is at distance |-11-(-3)|=8\neq2\sqrt5 from C, so it is not a tangent. Hence the two slopes give both tangents.
First tangent (m=2).
y+10=2(x+11)\;\Longrightarrow\;2x-y+12=0
Second tangent (m=\tfrac{2}{11}). Multiply by 11 to clear the fraction.
y+10=\dfrac{2}{11}(x+11)\;\Longrightarrow\;11y+110=2x+22\;\Longrightarrow\;2x-11y-88=0
**Point of contact on 2x-y+12=0.** Put y=2x+12 into the circle. A tangent meets the circle once, so we expect a perfect square.
x^2+(2x+12)^2+6x+8(2x+12)+5=0\;\Longrightarrow\;5x^2+70x+245=0\;\Longrightarrow\;(x+7)^2=0
So x=-7 and y=2(-7)+12=-2.
T_1(-7,\,-2)
**Point of contact on 2x-11y-88=0.** Put x=\dfrac{11y+88}{2} into the circle and multiply by 4.
(11y+88)^2+4y^2+12(11y+88)+32y+20=0\;\Longrightarrow\;125y^2+2100y+8820=0
Divide by 5; again a perfect square appears.
25y^2+420y+1764=0\;\Longrightarrow\;(5y+42)^2=0\;\Longrightarrow\;y=-\dfrac{42}{5}
Then x=\dfrac{11\left(-\frac{42}{5}\right)+88}{2}=\dfrac{-462+440}{10}=-\dfrac{11}{5}.
T_2\left(-\dfrac{11}{5},\,-\dfrac{42}{5}\right)
Check. Both contact points satisfy the circle, for example (-7)^2+(-2)^2+6(-7)+8(-2)+5=49+4-42-16+5=0.

\begin{aligned}&2x-y+12=0,\ \text{touching at }(-7,-2)\\ &2x-11y-88=0,\ \text{touching at }\left(-\tfrac{11}{5},-\tfrac{42}{5}\right)\end{aligned}

A quadrilateral ABCD is circumscribing a circle. Prove that \overline{AB}+\overline{CD}=\overline{AD}+\overline{BC}.

Solution

The circle (centre O) touches AB, BC, CD, DA at P, Q, R, S. The equal tangents AP and AS from A are drawn thick.

Key fact. The two tangents drawn to a circle from an outside point are equal in length (Section 7.6). For example from A: the radii OP and OS are perpendicular to the tangents, so by Pythagoras in the right triangles OPA and OSA,
\overline{AP}^{\,2}=\overline{OA}^{\,2}-r^2=\overline{AS}^{\,2}\;\Longrightarrow\;\overline{AP}=\overline{AS}
Let the circle touch AB, BC, CD, DA at P, Q, R, S. Apply the key fact at each vertex.
\overline{AP}=\overline{AS},\quad \overline{BP}=\overline{BQ},\quad \overline{CQ}=\overline{CR},\quad \overline{DR}=\overline{DS}
P lies on side AB and R lies on side CD, so split these two sides at the points of contact.
\overline{AB}+\overline{CD}=(\overline{AP}+\overline{PB})+(\overline{CR}+\overline{RD})
Replace each tangent length by its equal from the second step.
=\overline{AS}+\overline{BQ}+\overline{CQ}+\overline{DS}
Group the pieces that make up sides AD and BC (S lies on AD, Q lies on BC).
=(\overline{AS}+\overline{SD})+(\overline{BQ}+\overline{QC})=\overline{AD}+\overline{BC}

\overline{AB}+\overline{CD}=\overline{AD}+\overline{BC}\quad\blacksquare

Two tangents are drawn from a point P(6,1) lying outside the circle x^2+y^2-8x-2y+14=0. Find the area of the shaded region.

In the book's figure the tangents touch the circle at A and B, C is the centre and CP meets the circle at D. The shaded region lies between the two tangents and the arc ADB.

Solution

Centre and radius. Compare with x^2+y^2+2gx+2fy+c=0: g=-4, f=-1, c=14.
C(4,1),\qquad r=\sqrt{16+1-14}=\sqrt3
Length of each tangent (Section 7.6).
|PA|=|PB|=\sqrt{6^2+1^2-8(6)-2(1)+14}=\sqrt{36+1-48-2+14}=\sqrt1=1
Distance from the centre.
|CP|=\sqrt{(6-4)^2+(1-1)^2}=2
Angle at the centre. The radius is perpendicular to the tangent, so triangle CAP has a right angle at A.
\cos\angle ACP=\dfrac{|CA|}{|CP|}=\dfrac{\sqrt3}{2}\;\Longrightarrow\;\angle ACP=30^\circ=\dfrac{\pi}{6}
By symmetry about CP, \angle BCP=\dfrac{\pi}{6} too.
\angle ACB=\theta=\dfrac{\pi}{3}
**Area of the kite CAPB.** It is made of two equal right triangles with legs r and |PA|.
2\times\dfrac12|CA|\,|PA|=\sqrt3\times1=\sqrt3
**Area of the sector CADB**, with the angle in radians.
\dfrac12r^2\theta=\dfrac12(3)\left(\dfrac{\pi}{3}\right)=\dfrac{\pi}{2}
Shaded area. It is the kite minus the sector.
\text{Area}=\sqrt3-\dfrac{\pi}{2}\approx1.732-1.571=0.161\ \text{square units}

Shaded region = kite CAPB minus sector CADB.

\text{Shaded area}=\sqrt3-\dfrac{\pi}{2}\approx0.16\ \text{square units}

From a point P lying outside the circle a tangent and secant lines are drawn. Prove that |\overline{AP}|^2=|\overline{PB}|\cdot|\overline{PD}|.

Solution

PA is the tangent; the secant cuts the circle at B (nearer to P) and D. M is the foot of the perpendicular from the centre C to BD.

Let C be the centre and r the radius. Draw CM\perp BD. The perpendicular from the centre to a chord bisects the chord, so M is the midpoint of BD.
|MB|=|MD|
B lies between P and M, and D lies beyond M, so write both secant lengths using |PM|.
|PB|=|PM|-|MB|,\qquad |PD|=|PM|+|MD|=|PM|+|MB|
Multiply; the product is a difference of two squares.
|PB|\cdot|PD|=(|PM|-|MB|)(|PM|+|MB|)=|PM|^2-|MB|^2
Triangles CMP and CMB are right-angled at M. Apply Pythagoras to each (note |CB|=r).
|PM|^2=|CP|^2-|CM|^2,\qquad |MB|^2=r^2-|CM|^2
Substitute; the |CM|^2 terms cancel.
|PB|\cdot|PD|=\left(|CP|^2-|CM|^2\right)-\left(r^2-|CM|^2\right)=|CP|^2-r^2
The radius CA is perpendicular to the tangent PA, so triangle CAP is right-angled at A.
|AP|^2=|CP|^2-r^2
The two results have the same right-hand side.
|AP|^2=|PB|\cdot|PD|

|\overline{AP}|^2=|\overline{PB}|\cdot|\overline{PD}|\quad\blacksquare

Exercise 7.4p. 199

Parabola: elements from the equation (Example 12), equations from given data (Examples 11, 13 to 15), and the vertex from a horizontal or vertical tangent.

Find focus, vertex, axis of symmetry, directrix, length of latus rectum, end points of the latus rectum of parabola with the given equations. Also draw the graph of the parabola.

Solution

Method (Example 12). Complete the square to reach one of (y-k)^2=\pm4a(x-h) or (x-h)^2=\pm4a(y-k), put X=x-h, Y=y-k, and read the elements from the standard parabola. Finally return to x and y using x=X+h, y=Y+k.

In every case the vertex is (X,Y)=(0,0) and the length of the latus rectum is 4a (a>0).
Standard form	Focus	Directrix	Ends of latus rectum	Opens
Y^2=4aX	(a,0)	X=-a	(a,\pm2a)	right
Y^2=-4aX	(-a,0)	X=a	(-a,\pm2a)	left
X^2=4aY	(0,a)	Y=-a	(\pm2a,a)	up
X^2=-4aY	(0,-a)	Y=a	(\pm2a,-a)	down

(i) y^2=6x

Compare with the standard form y^2=4ax. The parabola opens to the right because y^2 equals a positive multiple of x.
4a=6\;\Longrightarrow\;a=\dfrac32
Vertex. The vertex of y^2=4ax is the origin.
V(0,\,0)
Focus. For y^2=4ax the focus is \left(a,\ 0\right).
F\left(\dfrac{3}{2},\ 0\right)
Axis of symmetry. The line through the vertex and the focus.
y=0
Directrix. It is on the other side of the vertex, at the same distance a.
x=- \dfrac{3}{2}
Length of latus rectum.
4a=4\left(\dfrac{3}{2}\right)=6
Ends of the latus rectum. The latus rectum passes through the focus, perpendicular to the axis, with half-length 2a=3 on each side.
\left(\dfrac{3}{2},\ 3\right),\qquad \left(\dfrac{3}{2},\ -3\right)
Graph. Plot V, F and the two ends of the latus rectum, draw the directrix as a dashed line, and join the three points of the parabola with a smooth curve that opens to the right, away from the directrix.

Dashed: directrix x=- \tfrac{3}{2}. Dotted: axis y=0. The short segment through F is the latus rectum.

\begin{aligned}&V\left(0,\ 0\right),\quad F\left(\tfrac{3}{2},\ 0\right)\\ &\text{axis } y=0\\ &\text{directrix } x=- \tfrac{3}{2}\\ &\text{latus rectum}=6\\ &\text{ends } \left(\tfrac{3}{2},\ 3\right),\ \left(\tfrac{3}{2},\ -3\right)\end{aligned}

(ii) y^2=-\dfrac32x

Compare with y^2=-4ax. It opens to the left because y^2 equals a negative multiple of x.
4a=\dfrac32\;\Longrightarrow\;a=\dfrac38
Vertex. The vertex of y^2=-4ax is the origin.
V(0,\,0)
Focus. For y^2=-4ax the focus is \left(-a,\ 0\right).
F\left(- \dfrac{3}{8},\ 0\right)
Axis of symmetry. The line through the vertex and the focus.
y=0
Directrix. It is on the other side of the vertex, at the same distance a.
x=\dfrac{3}{8}
Length of latus rectum.
4a=4\left(\dfrac{3}{8}\right)=\dfrac{3}{2}
Ends of the latus rectum. The latus rectum passes through the focus, perpendicular to the axis, with half-length 2a=\tfrac{3}{4} on each side.
\left(- \dfrac{3}{8},\ \dfrac{3}{4}\right),\qquad \left(- \dfrac{3}{8},\ - \dfrac{3}{4}\right)
Graph. Plot V, F and the two ends of the latus rectum, draw the directrix as a dashed line, and join the three points of the parabola with a smooth curve that opens to the left, away from the directrix.

Dashed: directrix x=\tfrac{3}{8}. Dotted: axis y=0. The short segment through F is the latus rectum.

\begin{aligned}&V\left(0,\ 0\right),\quad F\left(- \tfrac{3}{8},\ 0\right)\\ &\text{axis } y=0\\ &\text{directrix } x=\tfrac{3}{8}\\ &\text{latus rectum}=\tfrac{3}{2}\\ &\text{ends } \left(- \tfrac{3}{8},\ \tfrac{3}{4}\right),\ \left(- \tfrac{3}{8},\ - \tfrac{3}{4}\right)\end{aligned}

(iii) x^2=24y

Compare with x^2=4ay. It opens upwards because x^2 equals a positive multiple of y.
4a=24\;\Longrightarrow\;a=6
Vertex. The vertex of x^2=4ay is the origin.
V(0,\,0)
Focus. For x^2=4ay the focus is \left(0,\ a\right).
F\left(0,\ 6\right)
Axis of symmetry. The line through the vertex and the focus.
x=0
Directrix. It is on the other side of the vertex, at the same distance a.
y=-6
Length of latus rectum.
4a=4\left(6\right)=24
Ends of the latus rectum. The latus rectum passes through the focus, perpendicular to the axis, with half-length 2a=12 on each side.
\left(12,\ 6\right),\qquad \left(-12,\ 6\right)
Graph. Plot V, F and the two ends of the latus rectum, draw the directrix as a dashed line, and join the three points of the parabola with a smooth curve that opens upwards, away from the directrix.

Dashed: directrix y=-6. Dotted: axis x=0. The short segment through F is the latus rectum.

\begin{aligned}&V\left(0,\ 0\right),\quad F\left(0,\ 6\right)\\ &\text{axis } x=0\\ &\text{directrix } y=-6\\ &\text{latus rectum}=24\\ &\text{ends } \left(12,\ 6\right),\ \left(-12,\ 6\right)\end{aligned}

(iv) x^2=-5y

Compare with x^2=-4ay. It opens downwards because x^2 equals a negative multiple of y.
4a=5\;\Longrightarrow\;a=\dfrac54
Vertex. The vertex of x^2=-4ay is the origin.
V(0,\,0)
Focus. For x^2=-4ay the focus is \left(0,\ -a\right).
F\left(0,\ - \dfrac{5}{4}\right)
Axis of symmetry. The line through the vertex and the focus.
x=0
Directrix. It is on the other side of the vertex, at the same distance a.
y=\dfrac{5}{4}
Length of latus rectum.
4a=4\left(\dfrac{5}{4}\right)=5
Ends of the latus rectum. The latus rectum passes through the focus, perpendicular to the axis, with half-length 2a=\tfrac{5}{2} on each side.
\left(\dfrac{5}{2},\ - \dfrac{5}{4}\right),\qquad \left(- \dfrac{5}{2},\ - \dfrac{5}{4}\right)
Graph. Plot V, F and the two ends of the latus rectum, draw the directrix as a dashed line, and join the three points of the parabola with a smooth curve that opens downwards, away from the directrix.

Dashed: directrix y=\tfrac{5}{4}. Dotted: axis x=0. The short segment through F is the latus rectum.

\begin{aligned}&V\left(0,\ 0\right),\quad F\left(0,\ - \tfrac{5}{4}\right)\\ &\text{axis } x=0\\ &\text{directrix } y=\tfrac{5}{4}\\ &\text{latus rectum}=5\\ &\text{ends } \left(\tfrac{5}{2},\ - \tfrac{5}{4}\right),\ \left(- \tfrac{5}{2},\ - \tfrac{5}{4}\right)\end{aligned}

(v) y^2-2y-12x-71=0

Keep the y terms on the left and move the other terms to the right.
y^2-2y=12x+71
Complete the square: add \left(\tfrac{-2}{2}\right)^2=1 to both sides.
y^2-2y+1=12x+72\;\Longrightarrow\;(y-1)^2=12(x+6)
Put X=x+6 and Y=y-1. The equation becomes Y^2=12X, which is Y^2=4aX. It opens to the right.
Y^2=12X=4(3)X\;\Longrightarrow\;a=3
Vertex. The vertex is where X=0 and Y=0.
x+6=0,\ \ y-1=0\;\Longrightarrow\;V\left(-6,\ 1\right)
Focus. For Y^2=4aX the focus is (X,Y)=\left(a,\ 0\right)=\left(3,\ 0\right).
x+6=3,\ \ y-1=0\;\Longrightarrow\;F\left(-3,\ 1\right)
Axis of symmetry. It is the line Y=0 through the vertex and the focus.
y-1=0\;\Longrightarrow\;y=1
Directrix. For Y^2=4aX it is X=-a.
x+6=-3\;\Longrightarrow\;x=-9
Length of latus rectum.
4a=4\left(3\right)=12
Ends of the latus rectum. They are (X,Y)=(a,\pm2a): the focus moved 2a=6 each way along the latus rectum. Add h to X and k to Y.
\left(-3,\ 7\right),\qquad \left(-3,\ -5\right)
Graph. Plot V, F and the two ends of the latus rectum, draw the directrix as a dashed line, and join the three points of the parabola with a smooth curve that opens to the right, away from the directrix.

Dashed: directrix x=-9. Dotted: axis y=1. The short segment through F is the latus rectum.

\begin{aligned}&V\left(-6,\ 1\right),\quad F\left(-3,\ 1\right)\\ &\text{axis } y=1\\ &\text{directrix } x=-9\\ &\text{latus rectum}=12\\ &\text{ends } \left(-3,\ 7\right),\ \left(-3,\ -5\right)\end{aligned}

(vi) 3x^2+42x+y+149=0

Keep the x terms on the left.
3x^2+42x=-y-149
Take out the coefficient 3 of x^2.
3\left(x^2+14x\right)=-y-149
Complete the square: adding \left(\tfrac{14}{2}\right)^2=49 inside the bracket adds 3\times49=147 to the left, so add 147 to the right as well.
3\left(x^2+14x+49\right)=-y-149+147\;\Longrightarrow\;3(x+7)^2=-(y+2)
Divide by 3.
(x+7)^2=-\dfrac13(y+2)
Put X=x+7 and Y=y+2. The equation becomes X^2=-\tfrac13Y, which is X^2=-4aY. It opens downwards.
X^2=-\dfrac13Y=-4\left(\dfrac1{12}\right)Y\;\Longrightarrow\;a=\dfrac1{12}
Vertex. The vertex is where X=0 and Y=0.
x+7=0,\ \ y+2=0\;\Longrightarrow\;V\left(-7,\ -2\right)
Focus. For X^2=-4aY the focus is (X,Y)=\left(0,\ -a\right)=\left(0,\ - \dfrac{1}{12}\right).
x+7=0,\ \ y+2=- \dfrac{1}{12}\;\Longrightarrow\;F\left(-7,\ - \dfrac{25}{12}\right)
Axis of symmetry. It is the line X=0 through the vertex and the focus.
x+7=0\;\Longrightarrow\;x=-7
Directrix. For X^2=-4aY it is Y=a.
y+2=\dfrac{1}{12}\;\Longrightarrow\;y=- \dfrac{23}{12}
Length of latus rectum.
4a=4\left(\dfrac{1}{12}\right)=\dfrac{1}{3}
Ends of the latus rectum. They are (X,Y)=(\pm2a,-a): the focus moved 2a=\tfrac{1}{6} each way along the latus rectum. Add h to X and k to Y.
\left(- \dfrac{41}{6},\ - \dfrac{25}{12}\right),\qquad \left(- \dfrac{43}{6},\ - \dfrac{25}{12}\right)
Graph. Plot V, F and the two ends of the latus rectum, draw the directrix as a dashed line, and join the three points of the parabola with a smooth curve that opens downwards, away from the directrix.

Dashed: directrix y=- \tfrac{23}{12}. Dotted: axis x=-7. The short segment through F is the latus rectum. This parabola is narrow (a=\tfrac1{12}), so the window is zoomed in: V(-7,-2), F\left(-7,-\tfrac{25}{12}\right), and the ends of the latus rectum are \left(-\tfrac{41}{6},-\tfrac{25}{12}\right) and \left(-\tfrac{43}{6},-\tfrac{25}{12}\right).

\begin{aligned}&V\left(-7,\ -2\right),\quad F\left(-7,\ - \tfrac{25}{12}\right)\\ &\text{axis } x=-7\\ &\text{directrix } y=- \tfrac{23}{12}\\ &\text{latus rectum}=\tfrac{1}{3}\\ &\text{ends } \left(- \tfrac{41}{6},\ - \tfrac{25}{12}\right),\ \left(- \tfrac{43}{6},\ - \tfrac{25}{12}\right)\end{aligned}

(vii) 4y^2+4y=15-32x

The y terms are on the left. Take out the coefficient 4 of y^2.
4\left(y^2+y\right)=15-32x
Complete the square: adding \left(\tfrac12\right)^2=\tfrac14 inside the bracket adds 4\times\tfrac14=1 to the left, so add 1 to the right.
4\left(y^2+y+\tfrac14\right)=16-32x\;\Longrightarrow\;4\left(y+\tfrac12\right)^2=-32\left(x-\tfrac12\right)
Divide by 4.
\left(y+\dfrac12\right)^2=-8\left(x-\dfrac12\right)
Put X=x-\tfrac12 and Y=y+\tfrac12. The equation becomes Y^2=-8X, which is Y^2=-4aX. It opens to the left.
Y^2=-8X=-4(2)X\;\Longrightarrow\;a=2
Vertex. The vertex is where X=0 and Y=0.
x-\dfrac{1}{2}=0,\ \ y+\dfrac{1}{2}=0\;\Longrightarrow\;V\left(\dfrac{1}{2},\ - \dfrac{1}{2}\right)
Focus. For Y^2=-4aX the focus is (X,Y)=\left(-a,\ 0\right)=\left(-2,\ 0\right).
x-\dfrac{1}{2}=-2,\ \ y+\dfrac{1}{2}=0\;\Longrightarrow\;F\left(- \dfrac{3}{2},\ - \dfrac{1}{2}\right)
Axis of symmetry. It is the line Y=0 through the vertex and the focus.
y+\dfrac{1}{2}=0\;\Longrightarrow\;y=- \dfrac{1}{2}
Directrix. For Y^2=-4aX it is X=a.
x-\dfrac{1}{2}=2\;\Longrightarrow\;x=\dfrac{5}{2}
Length of latus rectum.
4a=4\left(2\right)=8
Ends of the latus rectum. They are (X,Y)=(-a,\pm2a): the focus moved 2a=4 each way along the latus rectum. Add h to X and k to Y.
\left(- \dfrac{3}{2},\ \dfrac{7}{2}\right),\qquad \left(- \dfrac{3}{2},\ - \dfrac{9}{2}\right)
Graph. Plot V, F and the two ends of the latus rectum, draw the directrix as a dashed line, and join the three points of the parabola with a smooth curve that opens to the left, away from the directrix.

Dashed: directrix x=\tfrac{5}{2}. Dotted: axis y=- \tfrac{1}{2}. The short segment through F is the latus rectum.

\begin{aligned}&V\left(\tfrac{1}{2},\ - \tfrac{1}{2}\right),\quad F\left(- \tfrac{3}{2},\ - \tfrac{1}{2}\right)\\ &\text{axis } y=- \tfrac{1}{2}\\ &\text{directrix } x=\tfrac{5}{2}\\ &\text{latus rectum}=8\\ &\text{ends } \left(- \tfrac{3}{2},\ \tfrac{7}{2}\right),\ \left(- \tfrac{3}{2},\ - \tfrac{9}{2}\right)\end{aligned}

(viii) 9x^2-6x=108y+26

The x terms are on the left. Take out the coefficient 9 of x^2.
9\left(x^2-\dfrac23x\right)=108y+26
Complete the square: adding \left(\tfrac13\right)^2=\tfrac19 inside the bracket adds 9\times\tfrac19=1 to the left, so add 1 to the right.
9\left(x-\dfrac13\right)^2=108y+27=108\left(y+\dfrac14\right)
Divide by 9.
\left(x-\dfrac13\right)^2=12\left(y+\dfrac14\right)
Put X=x-\tfrac13 and Y=y+\tfrac14. The equation becomes X^2=12Y, which is X^2=4aY. It opens upwards.
X^2=12Y=4(3)Y\;\Longrightarrow\;a=3
Vertex. The vertex is where X=0 and Y=0.
x-\dfrac{1}{3}=0,\ \ y+\dfrac{1}{4}=0\;\Longrightarrow\;V\left(\dfrac{1}{3},\ - \dfrac{1}{4}\right)
Focus. For X^2=4aY the focus is (X,Y)=\left(0,\ a\right)=\left(0,\ 3\right).
x-\dfrac{1}{3}=0,\ \ y+\dfrac{1}{4}=3\;\Longrightarrow\;F\left(\dfrac{1}{3},\ \dfrac{11}{4}\right)
Axis of symmetry. It is the line X=0 through the vertex and the focus.
x-\dfrac{1}{3}=0\;\Longrightarrow\;x=\dfrac{1}{3}
Directrix. For X^2=4aY it is Y=-a.
y+\dfrac{1}{4}=-3\;\Longrightarrow\;y=- \dfrac{13}{4}
Length of latus rectum.
4a=4\left(3\right)=12
Ends of the latus rectum. They are (X,Y)=(\pm2a,a): the focus moved 2a=6 each way along the latus rectum. Add h to X and k to Y.
\left(\dfrac{19}{3},\ \dfrac{11}{4}\right),\qquad \left(- \dfrac{17}{3},\ \dfrac{11}{4}\right)
Graph. Plot V, F and the two ends of the latus rectum, draw the directrix as a dashed line, and join the three points of the parabola with a smooth curve that opens upwards, away from the directrix.

Dashed: directrix y=- \tfrac{13}{4}. Dotted: axis x=\tfrac{1}{3}. The short segment through F is the latus rectum.

\begin{aligned}&V\left(\tfrac{1}{3},\ - \tfrac{1}{4}\right),\quad F\left(\tfrac{1}{3},\ \tfrac{11}{4}\right)\\ &\text{axis } x=\tfrac{1}{3}\\ &\text{directrix } y=- \tfrac{13}{4}\\ &\text{latus rectum}=12\\ &\text{ends } \left(\tfrac{19}{3},\ \tfrac{11}{4}\right),\ \left(- \tfrac{17}{3},\ \tfrac{11}{4}\right)\end{aligned}

Find the equation of the parabola in each of the following.

Solution

Plan. Decide the direction of the axis first: a focus and vertex on a horizontal line, or a vertical directrix, give (y-k)^2=\pm4a(x-h); a vertical axis or a horizontal directrix give (x-h)^2=\pm4a(y-k). Then a is the distance from the vertex to the focus (or to the directrix), and the parabola opens towards the focus, away from the directrix.

(i) Vertex at origin and focus \left(0,-\dfrac{1}{32}\right)

The focus is on the y-axis, below the vertex. So the axis is the y-axis and the parabola opens downwards: its equation has the form x^2=-4ay with a>0.
a is the distance from the vertex to the focus.
a=\left|0-\left(-\dfrac{1}{32}\right)\right|=\dfrac{1}{32}
Substitute into x^2=-4ay.
x^2=-4\left(\dfrac{1}{32}\right)y=-\dfrac18y\;\Longrightarrow\;8x^2=-y

A very narrow parabola: the focus is only \tfrac1{32} below the vertex. Dashed: directrix y=\tfrac1{32}.

8x^2+y=0\quad\left(x^2=-\tfrac18y\right)

(ii) Vertex (-8,-9) and focus \left(-\dfrac{31}{4},-9\right)

V and F have the same y-coordinate -9, so the axis is the horizontal line y=-9. The focus is to the right of the vertex (-\tfrac{31}{4}>-8), so the parabola opens to the right.
(y-k)^2=4a(x-h),\qquad (h,k)=(-8,-9)
a is the distance |VF|.
a=\left|-\dfrac{31}{4}-(-8)\right|=\dfrac14
Substitute h, k and a.
(y+9)^2=4\left(\dfrac14\right)(x+8)\;\Longrightarrow\;(y+9)^2=x+8
Expand to get the general form.
y^2+18y+81=x+8\;\Longrightarrow\;y^2-x+18y+73=0

Dashed: directrix x=-\tfrac{33}{4}, the same distance \tfrac14 on the other side of V.

(y+9)^2=x+8\quad\text{or}\quad y^2-x+18y+73=0

(iii) Vertex (-6,-9) and directrix: x=-\dfrac{47}{8}

The directrix is a vertical line, so the axis is horizontal: the line y=-9 through the vertex. The directrix x=-\tfrac{47}{8}=-5.875 is to the right of the vertex and the focus is on the opposite side, so the parabola opens to the left.
(y-k)^2=-4a(x-h),\qquad (h,k)=(-6,-9)
a is the distance from the vertex to the directrix.
a=\left|-\dfrac{47}{8}-(-6)\right|=\dfrac18
Substitute h, k and a.
(y+9)^2=-4\left(\dfrac18\right)(x+6)=-\dfrac12(x+6)
Multiply by 2 and expand.
2\left(y^2+18y+81\right)=-x-6\;\Longrightarrow\;2y^2+x+36y+168=0
The focus is a=\tfrac18 to the left of the vertex, at F\left(-\tfrac{49}{8},-9\right).

(y+9)^2=-\tfrac12(x+6)\quad\text{or}\quad 2y^2+x+36y+168=0

(iv) Vertex (5,-1) and y-intercept: -\dfrac{27}{2}

A parabola with a vertical axis (a graph y=Ax^2+Bx+C) cuts the y-axis at exactly one point, so \"the y-intercept\" points to a vertical axis. Take (x-h)^2=4a(y-k) with (h,k)=(5,-1); a negative a will mean it opens downwards.
(x-5)^2=4a(y+1)
The y-intercept -\tfrac{27}{2} means the point \left(0,-\tfrac{27}{2}\right) is on the parabola. Substitute it.
(0-5)^2=4a\left(-\dfrac{27}{2}+1\right)\;\Longrightarrow\;25=4a\left(-\dfrac{25}{2}\right)=-50a
Solve for a.
a=-\dfrac12\;\Longrightarrow\;4a=-2
Substitute back. Since a<0 the parabola opens downwards.
(x-5)^2=-2(y+1)
Expand.
x^2-10x+25=-2y-2\;\Longrightarrow\;x^2-10x+2y+27=0
Check. Put x=0: 2y+27=0, so y=-\tfrac{27}{2}, as required.

(x-5)^2=-2(y+1)\quad\text{or}\quad x^2-10x+2y+27=0

(v) Focus \left(-\dfrac34,-1\right) and directrix y=\dfrac25

Let P(x,y) be any point on the parabola and M the foot of the perpendicular from P to the directrix y-\tfrac25=0. By the definition of a parabola (as in Example 11), |PF|=|PM|.
\sqrt{\left(x+\tfrac34\right)^2+(y+1)^2}=\dfrac{\left|y-\tfrac25\right|}{\sqrt{0^2+1^2}}
Square both sides.
\left(x+\dfrac34\right)^2+(y+1)^2=\left(y-\dfrac25\right)^2
Keep the x term on the left and expand the y terms; the y^2 terms cancel.
\left(x+\dfrac34\right)^2=y^2-\dfrac45y+\dfrac4{25}-y^2-2y-1=-\dfrac{14}{5}y-\dfrac{21}{25}
Take out -\tfrac{14}{5} to get the standard form.
\left(x+\dfrac34\right)^2=-\dfrac{14}{5}\left(y+\dfrac3{10}\right)
So the vertex is \left(-\tfrac34,-\tfrac3{10}\right), halfway between the focus (y=-1) and the directrix (y=\tfrac25), with 4a=\tfrac{14}{5}. It opens downwards because the focus is below the directrix.
For the general form, expand to x^2+\tfrac32x+\tfrac9{16}+\tfrac{14}{5}y+\tfrac{21}{25}=0 and multiply by 400.
400x^2+600x+1120y+561=0

\left(x+\tfrac34\right)^2=-\tfrac{14}{5}\left(y+\tfrac3{10}\right)\quad\text{or}\quad 400x^2+600x+1120y+561=0

(vi) Opens left or right with vertex (7,6) and passes through (-11,9).

Opening left or right means the axis is horizontal. With vertex (h,k)=(7,6) the equation is
(y-6)^2=4a(x-7)
The point (-11,9) lies on the parabola; substitute it.
(9-6)^2=4a(-11-7)\;\Longrightarrow\;9=-72a\;\Longrightarrow\;a=-\dfrac18
Substitute 4a=-\tfrac12. As a<0, the parabola opens to the left.
(y-6)^2=-\dfrac12(x-7)
Multiply by 2 and expand.
2\left(y^2-12y+36\right)=-x+7\;\Longrightarrow\;2y^2+x-24y+65=0
Check with (-11,9): 2(81)+(-11)-24(9)+65=162-11-216+65=0.

The focus \left(\tfrac{55}{8},6\right) is only \tfrac18 from the vertex, too close to show at this scale.

(y-6)^2=-\tfrac12(x-7)\quad\text{or}\quad 2y^2+x-24y+65=0

(vii) Opens up or down and passes through the points (11,15), (7,7) and (4,22).

Opening up or down means the axis is vertical, so the parabola is a graph y=Ax^2+Bx+C (Section 7.7.6) with three unknowns. Substitute the three points.
\begin{aligned}(11,15):&\quad 121A+11B+C=15\quad &(1)\\ (7,7):&\quad 49A+7B+C=7 &(2)\\ (4,22):&\quad 16A+4B+C=22 &(3)\end{aligned}
Subtract to remove C: (1) minus (2), and (2) minus (3).
\begin{aligned}72A+4B&=8\\ \Rightarrow\ 18A+B&=2&&(4)\\[4pt] 33A+3B&=-15\\ \Rightarrow\ 11A+B&=-5&&(5)\end{aligned}
(4) minus (5) gives A. Then (4) gives B and (2) gives C.
7A=7\;\Rightarrow\;A=1,\qquad B=2-18=-16,\qquad C=7-49+112=70
Write the equation.
y=x^2-16x+70
Complete the square to see the vertex: the parabola opens upwards with vertex (8,6) and 4a=1.
y-6=x^2-16x+64\;\Longrightarrow\;(x-8)^2=y-6
Check with (4,22): 16-64+70=22.

y=x^2-16x+70\quad\text{or}\quad (x-8)^2=y-6

(viii) Vertex (10,0); axis of symmetry: y=0; length of latus rectum =1; a<0

The axis y=0 is horizontal, so with vertex (h,k)=(10,0) the equation is
(y-0)^2=4a(x-10)
The length of the latus rectum is |4a|=1, and a<0, so 4a=-1: the parabola opens to the left.
4a=-1\quad\left(a=-\tfrac14\right)
Substitute.
y^2=-(x-10)\;\Longrightarrow\;y^2+x-10=0
Its focus is \tfrac14 to the left of the vertex, at \left(\tfrac{39}{4},0\right); the ends of the latus rectum are \left(\tfrac{39}{4},\pm\tfrac12\right), which are 1 unit apart.

y^2=-(x-10)\quad\text{or}\quad y^2+x-10=0

(ix) Vertex (4,2); axis of symmetry: x=4; length of latus rectum =\dfrac13; a>0

The axis x=4 is vertical, so with vertex (h,k)=(4,2) the equation is
(x-4)^2=4a(y-2)
The latus rectum has length 4a=\tfrac13 (positive, since a>0): the parabola opens upwards.
4a=\dfrac13
Substitute and expand.
(x-4)^2=\dfrac13(y-2)\;\Longrightarrow\;3\left(x^2-8x+16\right)=y-2\;\Longrightarrow\;3x^2-24x-y+50=0
Its focus is a=\tfrac1{12} above the vertex, at \left(4,\tfrac{25}{12}\right).

Zoomed in: the parabola is narrow because 4a=\tfrac13 is small. Dashed: directrix y=\tfrac{23}{12}.

(x-4)^2=\tfrac13(y-2)\quad\text{or}\quad 3x^2-24x-y+50=0

(x) Vertex at origin; opens left; distance between focus and vertex is \dfrac18 units.

Vertex at the origin and opening to the left give the form y^2=-4ax, where a is the distance between the focus and the vertex.
a=\dfrac18
Substitute.
y^2=-4\left(\dfrac18\right)x=-\dfrac12x\;\Longrightarrow\;2y^2+x=0
The focus is \left(-\tfrac18,0\right) and the directrix is x=\tfrac18.

y^2=-\tfrac12x\quad\text{or}\quad 2y^2+x=0

Find the equation of the parabola with the focus at (p\sin\theta,\,p\cos\theta) whose directrix is x\cos\theta+y\sin\theta=p.

Solution

Let P(x,y) be any point of the parabola and F(p\sin\theta,\,p\cos\theta) its focus. By definition, |PF| equals the perpendicular distance |PM| of P from the directrix x\cos\theta+y\sin\theta-p=0.
\sqrt{(x-p\sin\theta)^2+(y-p\cos\theta)^2}=\dfrac{|x\cos\theta+y\sin\theta-p|}{\sqrt{\cos^2\theta+\sin^2\theta}}
The denominator is \sqrt1=1. Square both sides.
(x-p\sin\theta)^2+(y-p\cos\theta)^2=(x\cos\theta+y\sin\theta-p)^2
Expand the left side and use \sin^2\theta+\cos^2\theta=1.
\text{LHS}=x^2+y^2-2px\sin\theta-2py\cos\theta+p^2
Expand the right side, the square of three terms.
\text{RHS}=x^2\cos^2\theta+y^2\sin^2\theta+p^2+2xy\sin\theta\cos\theta-2px\cos\theta-2py\sin\theta
Put LHS minus RHS equal to zero. p^2 cancels, x^2-x^2\cos^2\theta=x^2\sin^2\theta and y^2-y^2\sin^2\theta=y^2\cos^2\theta.
x^2\sin^2\theta-2xy\sin\theta\cos\theta+y^2\cos^2\theta+2px(\cos\theta-\sin\theta)-2py(\cos\theta-\sin\theta)=0
The first three terms are a perfect square, and the last two have the common factor 2p(\cos\theta-\sin\theta).
(x\sin\theta-y\cos\theta)^2+2p(\cos\theta-\sin\theta)(x-y)=0
Rearrange.
(x\sin\theta-y\cos\theta)^2=2p(\sin\theta-\cos\theta)(x-y)
**Check with \theta=90^\circ.** The focus becomes (p,0) and the directrix y=p. The result gives x^2=2p(x-y), that is (x-p)^2=-2p\left(y-\tfrac p2\right): a parabola opening downwards with vertex \left(p,\tfrac p2\right), halfway between the focus and the directrix, as it should be.

The special case \theta=90^\circ, p=2: focus (2,0), directrix y=2 (dashed).

(x\sin\theta-y\cos\theta)^2=2p(\sin\theta-\cos\theta)(x-y)

Find the coordinates of the vertex of each parabola by differentiating its equation both sides and then solving for horizontal (or vertical) tangent.

Solution

Idea. At the vertex the tangent is perpendicular to the axis. If the equation contains x^2 (vertical axis), the tangent at the vertex is horizontal: solve \dfrac{dy}{dx}=0. If it contains y^2 (horizontal axis), the tangent at the vertex is vertical: \dfrac{dy}{dx} does not exist there, because its denominator is zero.

(i) y=x^2-4x+10

The equation has x^2 but no y^2, so the axis is vertical and the tangent at the vertex is horizontal. Differentiate.
\dfrac{dy}{dx}=2x-4
A horizontal tangent has slope zero.
2x-4=0\;\Longrightarrow\;x=2
Find y from the equation.
y=(2)^2-4(2)+10=6
Check by completing the square: y=(x-2)^2+6, so the vertex is (2,6).

The tangent at the vertex is the horizontal line y=6.

V(2,\,6)

(ii) 4x^2+24x+39-3y=0

Axis vertical (x^2 term), so look for a horizontal tangent. Differentiate both sides with respect to x.
8x+24-3\dfrac{dy}{dx}=0\;\Longrightarrow\;\dfrac{dy}{dx}=\dfrac{8x+24}{3}
Horizontal tangent: slope zero.
\dfrac{8x+24}{3}=0\;\Longrightarrow\;x=-3
Substitute x=-3 in the equation.
4(9)+24(-3)+39-3y=0\;\Longrightarrow\;3y=36-72+39=3\;\Longrightarrow\;y=1
Check. 4(x+3)^2-36+39=3y gives 4(x+3)^2=3(y-1), vertex (-3,1).

V(-3,\,1)

(iii) y^2-10y+4x+28=0

The equation has y^2 but no x^2, so the axis is horizontal and the tangent at the vertex is vertical. Differentiate with respect to x.
2y\dfrac{dy}{dx}-10\dfrac{dy}{dx}+4=0\;\Longrightarrow\;\dfrac{dy}{dx}=\dfrac{-4}{2y-10}=\dfrac{-2}{y-5}
A vertical tangent has no finite slope, so the denominator must be zero.
y-5=0\;\Longrightarrow\;y=5
Substitute y=5 in the equation.
25-50+4x+28=0\;\Longrightarrow\;4x=-3\;\Longrightarrow\;x=-\dfrac34
Check. (y-5)^2=-4x-3=-4\left(x+\tfrac34\right), vertex \left(-\tfrac34,5\right).

V\left(-\tfrac34,\,5\right)

(iv) y^2-14y=-3x-45y

The equation has y^2 but no x^2, so the tangent at the vertex is vertical. Differentiate with respect to x.
2y\dfrac{dy}{dx}-14\dfrac{dy}{dx}=-3\;\Longrightarrow\;\dfrac{dy}{dx}=\dfrac{-3}{2y-14}
Vertical tangent: the denominator is zero.
2y-14=0\;\Longrightarrow\;y=7
Substitute y=7.
49-98=-3x-45\;\Longrightarrow\;-4=-3x\;\Longrightarrow\;x=\dfrac43
Check. (y-7)^2=-3x-45+49=-3\left(x-\tfrac43\right), vertex \left(\tfrac43,7\right).
As printed (-45y): the equation is y^2+31y+3x=0, so \dfrac{dy}{dx}=\dfrac{-3}{2y+31}. The denominator is zero at y=-\tfrac{31}{2}, and then 3x=-y^2-31y=\tfrac{961}{4}.
V\left(\dfrac{961}{12},\,-\dfrac{31}{2}\right)

V\left(\tfrac43,\,7\right)\qquad\left(\text{as printed: }V\left(\tfrac{961}{12},-\tfrac{31}{2}\right)\right)

Note. Part (iv) is printed as y^2-14y=-3x-45y; the last term is taken as -45 (a misprint). The vertex for the printed equation is also given.

Exercise 7.5p. 202

Line and parabola: substitute the line into the parabola and use the discriminant (two points, tangent, or no point). Tangent and normal at a point: slope from \dfrac{dy}{dx} (Example 17). Proofs use the point (at^2,2at) on y^2=4ax.

Find the points of intersections of the line x-2y+3=0 and the parabola y^2=8x+1. Also find the chord intercepted. Is the chord a focal chord?

Solution

From the line, x=2y-3. Substitute into the parabola to get an equation in y only.
y^2=8(2y-3)+1=16y-23\;\Longrightarrow\;y^2-16y+23=0
Solve by the quadratic formula.
y=\dfrac{16\pm\sqrt{256-92}}{2}=\dfrac{16\pm\sqrt{164}}{2}=\dfrac{16\pm2\sqrt{41}}{2}=8\pm\sqrt{41}
Find x=2y-3 for each value.
x=2\left(8\pm\sqrt{41}\right)-3=13\pm2\sqrt{41}
So the points of intersection are
\begin{aligned}A\left(13+2\sqrt{41},\ 8+\sqrt{41}\right)&\approx(25.81,\,14.40)\\ B\left(13-2\sqrt{41},\ 8-\sqrt{41}\right)&\approx(0.19,\,1.60)\end{aligned}
Length of the chord AB by the distance formula. The differences are x_A-x_B=4\sqrt{41} and y_A-y_B=2\sqrt{41}.
|AB|=\sqrt{\left(4\sqrt{41}\right)^2+\left(2\sqrt{41}\right)^2}=\sqrt{656+164}=\sqrt{820}=2\sqrt{205}\approx28.64
Is it a focal chord? Write the parabola in standard form to find its focus.
y^2=8\left(x+\dfrac18\right)\;\Longrightarrow\;Y^2=4(2)X\quad\text{with } X=x+\dfrac18,\ Y=y
So a=2 and the focus is (X,Y)=(2,0), that is x=2-\tfrac18, y=0.
F\left(\dfrac{15}{8},\,0\right)
A focal chord passes through the focus. Substitute F into the left side of the line.
\dfrac{15}{8}-2(0)+3=\dfrac{39}{8}\neq0
The focus is not on the line, so the chord AB is not a focal chord.

The chord AB (length about 28.6) passes well above the focus F.

\begin{aligned}&A\left(13+2\sqrt{41},\,8+\sqrt{41}\right),\quad B\left(13-2\sqrt{41},\,8-\sqrt{41}\right)\\ &|AB|=2\sqrt{205}\approx28.64;\ \text{not a focal chord}\end{aligned}

For what value(s) of c the line 2x+2y-c=0 never touches or intersects at different points of the parabola x^2-x+2y+3=0.

Solution

From the line, 2y=c-2x. The parabola contains 2y, so substitute directly.
x^2-x+(c-2x)+3=0\;\Longrightarrow\;x^2-3x+(c+3)=0
Each real root of this quadratic gives a common point of the line and the parabola. The discriminant decides how many there are.
\text{Disc}=(-3)^2-4(1)(c+3)=9-4c-12=-3-4c
Intersects at two different points: two distinct real roots, so Disc >0. Dividing by -4 reverses the inequality.
-3-4c>0\;\Longrightarrow\;-4c>3\;\Longrightarrow\;c<-\dfrac34
Never touches (no common point): no real roots, so Disc <0.
-3-4c<0\;\Longrightarrow\;c>-\dfrac34
The boundary value c=-\tfrac34 gives Disc =0: then the line is a tangent, touching the parabola at x=\tfrac32, y=-\tfrac{15}{8}.

Lines y=\tfrac c2-x for three values of c.

\text{two different points: } c<-\tfrac34;\qquad \text{never touches: } c>-\tfrac34\qquad(\text{tangent when } c=-\tfrac34)

Find the value of a so that the line ax-2y+3=0 is tangent to the parabola: y^2-2y+3x+7=0; a\neq0

Solution

The parabola is quadratic in y, so eliminate x. From the line (with a\neq0):
x=\dfrac{2y-3}{a}
Substitute into the parabola and multiply by a.
y^2-2y+\dfrac{3(2y-3)}{a}+7=0\;\Longrightarrow\;ay^2-2ay+6y-9+7a=0
Collect terms: a quadratic in y.
ay^2+(6-2a)y+(7a-9)=0
A tangent meets the parabola at one point only, so this quadratic must have equal roots: Disc =0.
(6-2a)^2-4a(7a-9)=0
Expand and simplify; divide by -12 at the end.
36-24a+4a^2-28a^2+36a=0\;\Longrightarrow\;-24a^2+12a+36=0\;\Longrightarrow\;2a^2-a-3=0
Factorise.
(2a-3)(a+1)=0\;\Longrightarrow\;a=\dfrac32\ \text{ or }\ a=-1
The tangent lines. For a=\tfrac32 the line is 3x-4y+6=0; the quadratic becomes \tfrac32(y+1)^2=0, so it touches at \left(-\tfrac{10}{3},-1\right). For a=-1 the line is x+2y-3=0; the quadratic becomes -(y-4)^2=0, so it touches at (-5,4).

a=\tfrac32\ \ (3x-4y+6=0)\qquad\text{or}\qquad a=-1\ \ (x+2y-3=0)

Prove that the line 3x+y-5=0 is tangent to the parabola y^2-2y+6x-6=0. Also find the point of contact.

Solution

From the line, 3x=5-y, so 6x=10-2y. Substitute into the parabola.
y^2-2y+(10-2y)-6=0\;\Longrightarrow\;y^2-4y+4=0
Find the discriminant of this quadratic in y.
\text{Disc}=(-4)^2-4(1)(4)=16-16=0
Disc =0 gives equal roots, so the line meets the parabola at exactly one point. Hence the line is a tangent.
Point of contact. Solve the quadratic, then use the line.
(y-2)^2=0\;\Longrightarrow\;y=2,\qquad x=\dfrac{5-y}{3}=\dfrac{5-2}{3}=1
Check with the slope. Differentiating the parabola: 2y\dfrac{dy}{dx}-2\dfrac{dy}{dx}+6=0, so \dfrac{dy}{dx}=\dfrac{-3}{y-1}=-3 at (1,2), the same as the slope of y=5-3x.

\text{The line is a tangent; point of contact } (1,\,2)

Find the equation of tangent and normal to the parabola 2y^2-3y+11x-16=0 at the point (1,-1).

Solution

Check that the point is on the parabola.
2(-1)^2-3(-1)+11(1)-16=2+3+11-16=0\ \checkmark
Differentiate with respect to x (as in Example 17).
4y\dfrac{dy}{dx}-3\dfrac{dy}{dx}+11=0\;\Longrightarrow\;\dfrac{dy}{dx}=\dfrac{-11}{4y-3}
Slope of the tangent at (1,-1).
m=\dfrac{-11}{4(-1)-3}=\dfrac{-11}{-7}=\dfrac{11}{7}
Tangent by the point-slope form.
y+1=\dfrac{11}{7}(x-1)\;\Longrightarrow\;7y+7=11x-11\;\Longrightarrow\;11x-7y-18=0
Normal. It is perpendicular to the tangent, so its slope is -\dfrac1m=-\dfrac{7}{11}.
y+1=-\dfrac{7}{11}(x-1)\;\Longrightarrow\;11y+11=-7x+7\;\Longrightarrow\;7x+11y+4=0

\text{Tangent: } 11x-7y-18=0,\qquad \text{Normal: } 7x+11y+4=0

Find the equation of tangent and normal to the parabola x^2-5x+2y+6=0 at the point where abscissa is 1.

Solution

The abscissa is the x-coordinate. Put x=1 in the equation to find the ordinate.
1-5+2y+6=0\;\Longrightarrow\;2y=-2\;\Longrightarrow\;y=-1;\qquad P(1,-1)
Differentiate with respect to x.
2x-5+2\dfrac{dy}{dx}=0\;\Longrightarrow\;\dfrac{dy}{dx}=\dfrac{5-2x}{2}
Slope of the tangent at P.
m=\dfrac{5-2(1)}{2}=\dfrac32
Tangent.
y+1=\dfrac32(x-1)\;\Longrightarrow\;2y+2=3x-3\;\Longrightarrow\;3x-2y-5=0
Normal, with slope -\dfrac1m=-\dfrac23.
y+1=-\dfrac23(x-1)\;\Longrightarrow\;3y+3=-2x+2\;\Longrightarrow\;2x+3y+1=0

\text{Tangent: } 3x-2y-5=0,\qquad \text{Normal: } 2x+3y+1=0

Find the equation of tangent and normal to the parabola y^2=18x at the end points of its latus rectum.

Solution

Compare with y^2=4ax.
4a=18\;\Longrightarrow\;a=\dfrac92,\qquad F\left(\dfrac92,\,0\right)
The ends of the latus rectum are (a,\pm2a).
L\left(\dfrac92,\,9\right),\qquad L'\left(\dfrac92,\,-9\right)
Differentiate y^2=18x to get the slope of the tangent at any point.
2y\dfrac{dy}{dx}=18\;\Longrightarrow\;\dfrac{dy}{dx}=\dfrac9y
**At L\left(\tfrac92,9\right)** the slope is \tfrac99=1.
\text{Tangent: } y-9=1\left(x-\dfrac92\right)\;\Longrightarrow\;2x-2y+9=0
The normal at L has slope -1.
\text{Normal: } y-9=-1\left(x-\dfrac92\right)\;\Longrightarrow\;2x+2y-27=0
**At L'\left(\tfrac92,-9\right)** the slope is \tfrac{9}{-9}=-1.
\text{Tangent: } y+9=-1\left(x-\dfrac92\right)\;\Longrightarrow\;2x+2y+9=0
The normal at L' has slope 1.
\text{Normal: } y+9=1\left(x-\dfrac92\right)\;\Longrightarrow\;2x-2y-27=0

Solid: the tangents at L and L', meeting on the directrix. Dashed: the normals, meeting on the axis at \left(\tfrac{27}{2},0\right).

\begin{aligned}\text{At }\left(\tfrac92,9\right):\ &\text{tangent } 2x-2y+9=0\\ &\text{normal } 2x+2y-27=0\\ \text{At }\left(\tfrac92,-9\right):\ &\text{tangent } 2x+2y+9=0\\ &\text{normal } 2x-2y-27=0\end{aligned}

Let P(at_1^2,2at_1) and Q(at_2^2,2at_2) be any two points on the parabola y^2=4ax. Prove that the chord joining P and Q is a focal chord if t_1t_2=-1.

Solution

Both points lie on the parabola, because (2at)^2=4a^2t^2=4a\left(at^2\right) for every t. The focus of y^2=4ax is S(a,0), and a focal chord is a chord through S.
Slope of PQ (with t_1\neq t_2). Factor the difference of squares and cancel a(t_1-t_2).
m=\dfrac{2at_1-2at_2}{at_1^2-at_2^2}=\dfrac{2a(t_1-t_2)}{a(t_1-t_2)(t_1+t_2)}=\dfrac{2}{t_1+t_2}
Equation of PQ through P (point-slope form).
y-2at_1=\dfrac{2}{t_1+t_2}\left(x-at_1^2\right)
Multiply by t_1+t_2; the terms -2at_1^2 on both sides cancel.
(t_1+t_2)y-2at_1^2-2at_1t_2=2x-2at_1^2\;\Longrightarrow\;2x-(t_1+t_2)y+2at_1t_2=0
Now use the given condition t_1t_2=-1.
2x-(t_1+t_2)y-2a=0
Substitute the focus S(a,0).
2(a)-(t_1+t_2)(0)-2a=0\ \checkmark
S lies on PQ, so PQ is a focal chord. (Conversely, if S is on PQ then 2a+2at_1t_2=0, which gives t_1t_2=-1.)

Illustration with a=1: t_1=2 and t_2=-\tfrac12 have t_1t_2=-1, and the chord PQ passes through the focus S.

t_1t_2=-1\;\Longrightarrow\;S(a,0)\ \text{lies on } PQ\quad\blacksquare

A tangent line is drawn to the parabola at any point P. Prove that the line segment of the tangent cut off between P and directrix subtends a right angle at the focus.

Solution

Take the parabola y^2=4ax, with focus S(a,0) and directrix x=-a. Any point on it can be written P(at^2,2at). Take t\neq0: at the vertex (t=0) the tangent x=0 is parallel to the directrix and never meets it.
Slope of the tangent at P: differentiate y^2=4ax.
2y\dfrac{dy}{dx}=4a\;\Longrightarrow\;\dfrac{dy}{dx}=\dfrac{2a}{y}=\dfrac{2a}{2at}=\dfrac1t
Equation of the tangent at P.
y-2at=\dfrac1t\left(x-at^2\right)\;\Longrightarrow\;ty-2at^2=x-at^2\;\Longrightarrow\;ty=x+at^2
It meets the directrix x=-a at the point Q.
ty=-a+at^2\;\Longrightarrow\;y=\dfrac{a\left(t^2-1\right)}{t};\qquad Q\left(-a,\ \dfrac{a\left(t^2-1\right)}{t}\right)
Slopes of SP and SQ (for t^2\neq1).
m_{SP}=\dfrac{2at-0}{at^2-a}=\dfrac{2t}{t^2-1},\qquad m_{SQ}=\dfrac{\frac{a(t^2-1)}{t}-0}{-a-a}=-\dfrac{t^2-1}{2t}
Multiply the slopes.
m_{SP}\cdot m_{SQ}=\dfrac{2t}{t^2-1}\cdot\left(-\dfrac{t^2-1}{2t}\right)=-1
So SP\perp SQ: the segment PQ of the tangent subtends a right angle at the focus.
\angle PSQ=90^\circ
If t^2=1, then P(a,\pm2a) is an end of the latus rectum: SP is vertical and Q=(-a,0), so SQ lies along the axis. The angle is again 90^\circ.

Illustration with a=1, t=2: SP has slope \tfrac43 and SQ has slope -\tfrac34, so \angle PSQ=90^\circ.

\angle PSQ=90^\circ\quad\blacksquare

Prove that tangents at the end points of any focal chord intersect at right angles on the directrix.

Solution

Let PQ be a focal chord of y^2=4ax, with P(at_1^2,2at_1) and Q(at_2^2,2at_2). By Question 8, a chord through the focus has
t_1t_2=-1
By Question 9, the tangent at the point with parameter t is ty=x+at^2, with slope \tfrac1t. So the tangents at P and Q are
t_1y=x+at_1^2\quad(1),\qquad t_2y=x+at_2^2\quad(2)
Product of their slopes. Neither parameter is zero, because t_1t_2=-1.
\dfrac1{t_1}\cdot\dfrac1{t_2}=\dfrac1{t_1t_2}=\dfrac{1}{-1}=-1
So the two tangents are perpendicular. Subtract (2) from (1) to find where they meet.
(t_1-t_2)y=a\left(t_1^2-t_2^2\right)\;\Longrightarrow\;y=a(t_1+t_2)
Put this value of y into (1).
x=t_1y-at_1^2=at_1(t_1+t_2)-at_1^2=at_1t_2=-a
The tangents meet at \left(-a,\ a(t_1+t_2)\right). Its x-coordinate is -a, so the point lies on the directrix x=-a.

Illustration with a=1, t_1=2, t_2=-\tfrac12: the tangents (slopes \tfrac12 and -2) meet at right angles on the directrix x=-1.

\begin{aligned}&\text{Tangents at } P,\,Q\\ &\text{meet at } \left(-a,\,a(t_1+t_2)\right)\\ &\text{on } x=-a, \text{ at right angles}\quad\blacksquare\end{aligned}

Prove that tangents to the parabola at any point P on the parabola make equal angle with line joining P and its focus and the line through P parallel to the axis of parabola. (Reflecting property of parabola).

Solution

Take y^2=4ax with focus S(a,0), axis y=0 and directrix x=-a, and a point P(at^2,2at) with t\neq0 (at the vertex the result is clear by symmetry). By Question 9 the tangent at P is ty=x+at^2.
The tangent meets the axis (y=0) at T.
0=x+at^2\;\Longrightarrow\;T\left(-at^2,\,0\right)
Distance ST along the axis.
|ST|=a-\left(-at^2\right)=a\left(1+t^2\right)
Distance SP by the distance formula.
|SP|=\sqrt{\left(at^2-a\right)^2+(2at)^2}=a\sqrt{t^4+2t^2+1}=a\sqrt{\left(t^2+1\right)^2}=a\left(t^2+1\right)
So |SP|=|ST|: triangle SPT is isosceles, and its base angles are equal.
\angle SPT=\angle STP
Let PM be the line through P parallel to the axis (M is on the directrix). Since PM\parallel TS and TP cuts both, the alternate angles are equal.
\angle MPT=\angle STP
From the last two steps, the tangent PT makes equal angles with PS and with PM.
\angle SPT=\angle MPT
Check with slopes. The tangent has slope m=\tfrac1t and SP has slope \tfrac{2t}{t^2-1}. For the angle \alpha between the tangent and SP, and the angle \beta between the tangent and the horizontal line PM:
\tan\alpha=\left|\dfrac{\frac{2t}{t^2-1}-\frac1t}{1+\frac{2t}{t^2-1}\cdot\frac1t}\right|=\left|\dfrac{\frac{t^2+1}{t(t^2-1)}}{\frac{t^2+1}{t^2-1}}\right|=\dfrac1{|t|}=\tan\beta

Illustration with a=1, t=2: |SP|=|ST|=5, so the tangent makes equal angles with PS and with PM (parallel to the axis).

\angle SPT=\angle MPT\quad\blacksquare

Prove that the semi latus rectum is a harmonic mean between the segments of any focal chord.

Solution

For y^2=4ax the latus rectum has length 4a, so the semi latus rectum is 2a. The focus S divides a focal chord PQ into the segments SP and SQ.
The harmonic mean H of two positive numbers u and v is defined by
\dfrac1H=\dfrac12\left(\dfrac1u+\dfrac1v\right),\quad\text{that is}\quad H=\dfrac{2uv}{u+v}
Let P(at_1^2,2at_1) and Q(at_2^2,2at_2). The chord passes through the focus, so t_1t_2=-1 (Question 8), that is t_2=-\dfrac1{t_1}.
By the definition of the parabola, the distance of a point from the focus equals its distance from the directrix x=-a, which is x+a.
|SP|=at_1^2+a=a\left(1+t_1^2\right)
Similarly for Q, using t_2^2=\dfrac1{t_1^2}.
|SQ|=a\left(1+t_2^2\right)=a\left(1+\dfrac1{t_1^2}\right)=\dfrac{a\left(t_1^2+1\right)}{t_1^2}
Add the reciprocals.
\dfrac1{|SP|}+\dfrac1{|SQ|}=\dfrac{1}{a\left(1+t_1^2\right)}+\dfrac{t_1^2}{a\left(1+t_1^2\right)}=\dfrac{1+t_1^2}{a\left(1+t_1^2\right)}=\dfrac1a
So the harmonic mean of |SP| and |SQ| is the semi latus rectum.
H=\dfrac{2}{\frac1{|SP|}+\frac1{|SQ|}}=\dfrac{2}{\,1/a\,}=2a
Example check: a=1, t_1=2 gives |SP|=5 and |SQ|=\tfrac54, and H=\dfrac{2(5)\left(\frac54\right)}{5+\frac54}=\dfrac{12.5}{6.25}=2=2a.

With a=1: |SP|=5, |SQ|=\tfrac54, and their harmonic mean is 2=|SL|, the semi latus rectum.

\dfrac{2\,|SP|\,|SQ|}{|SP|+|SQ|}=2a=\text{semi latus rectum}\quad\blacksquare

Exercise 7.6p. 207

Ellipse: elements from the equation (Example 19), the equation from given data, and the ends of the axes from horizontal and vertical tangents.

Find the centre, vertices, co-vertices, foci, eccentricity, length and equation of major axis, length and equation of minor axis, directrices, and length of latus rectums for the following equations of ellipse. Also draw the ellipse in each case

Solution

Method (Example 19). Complete the squares to reach \dfrac{(x-h)^2}{a^2}+\dfrac{(y-k)^2}{b^2}=1 (major axis horizontal) or \dfrac{(x-h)^2}{b^2}+\dfrac{(y-k)^2}{a^2}=1 (major axis vertical), where a>b. Then use c^2=a^2-b^2, e=\dfrac ca, directrices at distance \dfrac ae=\dfrac{a^2}c from the centre, and latus rectum \dfrac{2b^2}{a}.

Elements of an ellipse with centre (h,k), a>b>0 and c=\sqrt{a^2-b^2}.
Element	Major axis horizontal	Major axis vertical
Centre	(h,k)	(h,k)
Vertices	(h\pm a,\ k)	(h,\ k\pm a)
Co-vertices	(h,\ k\pm b)	(h\pm b,\ k)
Foci	(h\pm c,\ k)	(h,\ k\pm c)
Major axis	y=k, length 2a	x=h, length 2a
Minor axis	x=h, length 2b	y=k, length 2b
Directrices	x=h\pm\dfrac{a}{e}	y=k\pm\dfrac{a}{e}

(i) \dfrac{x^2}{9}+\dfrac{y^2}{16}=1

The equation is already in standard form with centre at the origin.
The larger denominator is under y^2, so the major axis is vertical. Compare with \dfrac{x^2}{b^2}+\dfrac{y^2}{a^2}=1.
a^2=16,\ \ b^2=9\;\Longrightarrow\;a=4,\ \ b=3
Find c from c^2=a^2-b^2 (the distance from the centre to each focus).
c=\sqrt{16-9}=\sqrt{7}
Centre. The equation has no first-degree terms, so the centre is the origin.
C(0,\,0)
Vertices are a units below and above the centre, (h,\,k\pm a).
A_1\left(0,\ -4\right),\qquad A_2\left(0,\ 4\right)
Co-vertices are b units left and right of the centre, (h\pm b,\,k).
B_1\left(-3,\ 0\right),\qquad B_2\left(3,\ 0\right)
Foci are on the major axis, c units from the centre, (h,\,k\pm c).
F_1\left(0,\ - \sqrt{7}\right),\qquad F_2\left(0,\ \sqrt{7}\right)
Eccentricity.
e=\dfrac ca=\dfrac{\sqrt{7}}{4}=\dfrac{\sqrt{7}}{4}\approx0.661\ (<1)
Major axis: length 2a; it is the line through the foci.
2a=8,\qquad\text{equation } x=0
Minor axis: length 2b; it is the line through the centre perpendicular to the major axis.
2b=6,\qquad\text{equation } y=0
Directrices are perpendicular to the major axis at distance \dfrac ae=\dfrac{a^2}{c} from the centre.
\dfrac{a^2}{c}=\dfrac{16}{\sqrt{7}}=\dfrac{16 \sqrt{7}}{7}\;\Longrightarrow\;y=\pm\dfrac{16 \sqrt{7}}{7}
Length of each latus rectum.
\dfrac{2b^2}{a}=\dfrac{2\left(9\right)}{4}=\dfrac{9}{2}
Graph. Plot the centre, the two vertices and the two co-vertices, and join the four points with a smooth oval. Mark the foci on the major axis and draw the directrices as dashed lines.

Dotted: major axis. Dashed: directrices y=\pm\tfrac{16 \sqrt{7}}{7}. F_1, F_2 are the foci.

\begin{aligned}&C\left(0,\ 0\right)\\ &A\left(0,\ -4\right),\,A'\left(0,\ 4\right)\\ &B\left(-3,\ 0\right),\,B'\left(3,\ 0\right)\\ &F\left(0,\ - \sqrt{7}\right),\,F'\left(0,\ \sqrt{7}\right)\\ &e=\tfrac{\sqrt{7}}{4}\\ &\text{major axis } 2a=8,\ x=0\\ &\text{minor axis } 2b=6,\ y=0\\ &\text{directrices } y=\pm\tfrac{16 \sqrt{7}}{7}\\ &\text{latus rectum}=\tfrac{9}{2}\end{aligned}

(ii) 2x^2+3y^2=30

Divide both sides by 30 to make the right side 1.
\dfrac{2x^2}{30}+\dfrac{3y^2}{30}=1\;\Longrightarrow\;\dfrac{x^2}{15}+\dfrac{y^2}{10}=1
The larger denominator is under x^2, so the major axis is horizontal. Compare with \dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1.
a^2=15,\ \ b^2=10\;\Longrightarrow\;a=\sqrt{15},\ \ b=\sqrt{10}
Find c from c^2=a^2-b^2 (the distance from the centre to each focus).
c=\sqrt{15-10}=\sqrt{5}
Centre. The equation has no first-degree terms, so the centre is the origin.
C(0,\,0)
Vertices are a units left and right of the centre, (h\pm a,\,k).
A_1\left(- \sqrt{15},\ 0\right),\qquad A_2\left(\sqrt{15},\ 0\right)
Co-vertices are b units below and above the centre, (h,\,k\pm b).
B_1\left(0,\ - \sqrt{10}\right),\qquad B_2\left(0,\ \sqrt{10}\right)
Foci are on the major axis, c units from the centre, (h\pm c,\,k).
F_1\left(- \sqrt{5},\ 0\right),\qquad F_2\left(\sqrt{5},\ 0\right)
Eccentricity.
e=\dfrac ca=\dfrac{\sqrt{5}}{\sqrt{15}}=\dfrac{\sqrt{3}}{3}\approx0.577\ (<1)
Major axis: length 2a; it is the line through the foci.
2a=2 \sqrt{15},\qquad\text{equation } y=0
Minor axis: length 2b; it is the line through the centre perpendicular to the major axis.
2b=2 \sqrt{10},\qquad\text{equation } x=0
Directrices are perpendicular to the major axis at distance \dfrac ae=\dfrac{a^2}{c} from the centre.
\dfrac{a^2}{c}=\dfrac{15}{\sqrt{5}}=3 \sqrt{5}\;\Longrightarrow\;x=\pm3 \sqrt{5}
Length of each latus rectum.
\dfrac{2b^2}{a}=\dfrac{2\left(10\right)}{\sqrt{15}}=\dfrac{4 \sqrt{15}}{3}\approx5.16
Graph. Plot the centre, the two vertices and the two co-vertices, and join the four points with a smooth oval. Mark the foci on the major axis and draw the directrices as dashed lines.

Dotted: major axis. Dashed: directrices x=\pm3 \sqrt{5}. F_1, F_2 are the foci.

\begin{aligned}&C\left(0,\ 0\right)\\ &A\left(- \sqrt{15},\ 0\right),\,A'\left(\sqrt{15},\ 0\right)\\ &B\left(0,\ - \sqrt{10}\right),\,B'\left(0,\ \sqrt{10}\right)\\ &F\left(- \sqrt{5},\ 0\right),\,F'\left(\sqrt{5},\ 0\right)\\ &e=\tfrac{\sqrt{3}}{3}\\ &\text{major axis } 2a=2 \sqrt{15},\ y=0\\ &\text{minor axis } 2b=2 \sqrt{10},\ x=0\\ &\text{directrices } x=\pm3 \sqrt{5}\\ &\text{latus rectum}=\tfrac{4 \sqrt{15}}{3}\end{aligned}

(iii) \dfrac{x^2}{49}+\dfrac{(y-3)^2}{64}=1

Put X=x and Y=y-3 (a shift of the centre to (0,3)).
\dfrac{X^2}{49}+\dfrac{Y^2}{64}=1
The larger denominator is under Y^2, so the major axis is vertical. Compare with \dfrac{X^2}{b^2}+\dfrac{Y^2}{a^2}=1.
a^2=64,\ \ b^2=49\;\Longrightarrow\;a=8,\ \ b=7
Find c from c^2=a^2-b^2 (the distance from the centre to each focus).
c=\sqrt{64-49}=\sqrt{15}
Centre. It is where X=0 and Y=0.
C\left(0,\ 3\right)
Vertices are a units below and above the centre, (h,\,k\pm a).
A_1\left(0,\ -5\right),\qquad A_2\left(0,\ 11\right)
Co-vertices are b units left and right of the centre, (h\pm b,\,k).
B_1\left(-7,\ 3\right),\qquad B_2\left(7,\ 3\right)
Foci are on the major axis, c units from the centre, (h,\,k\pm c).
F_1\left(0,\ 3 - \sqrt{15}\right),\qquad F_2\left(0,\ 3 + \sqrt{15}\right)
Eccentricity.
e=\dfrac ca=\dfrac{\sqrt{15}}{8}=\dfrac{\sqrt{15}}{8}\approx0.484\ (<1)
Major axis: length 2a; it is the line through the foci.
2a=16,\qquad\text{equation } x=0
Minor axis: length 2b; it is the line through the centre perpendicular to the major axis.
2b=14,\qquad\text{equation } y=3
Directrices are perpendicular to the major axis at distance \dfrac ae=\dfrac{a^2}{c} from the centre.
\dfrac{a^2}{c}=\dfrac{64}{\sqrt{15}}=\dfrac{64 \sqrt{15}}{15}\;\Longrightarrow\;y=3\pm\dfrac{64 \sqrt{15}}{15}
Length of each latus rectum.
\dfrac{2b^2}{a}=\dfrac{2\left(49\right)}{8}=\dfrac{49}{4}
Graph. Plot the centre, the two vertices and the two co-vertices, and join the four points with a smooth oval. Mark the foci on the major axis and draw the directrices as dashed lines.

Dotted: major axis. Dashed: directrices y=3\pm\tfrac{64 \sqrt{15}}{15}. F_1, F_2 are the foci.

\begin{aligned}&C\left(0,\ 3\right)\\ &A\left(0,\ -5\right),\,A'\left(0,\ 11\right)\\ &B\left(-7,\ 3\right),\,B'\left(7,\ 3\right)\\ &F\left(0,\ 3 - \sqrt{15}\right),\,F'\left(0,\ 3 + \sqrt{15}\right)\\ &e=\tfrac{\sqrt{15}}{8}\\ &\text{major axis } 2a=16,\ x=0\\ &\text{minor axis } 2b=14,\ y=3\\ &\text{directrices } y=3\pm\tfrac{64 \sqrt{15}}{15}\\ &\text{latus rectum}=\tfrac{49}{4}\end{aligned}

(iv) x^2+9y^2+6x-90y+225=0

Group the x terms and the y terms, and take out the coefficient 9 of y^2.
\left(x^2+6x\right)+9\left(y^2-10y\right)=-225
Complete both squares. Adding 9 in the first bracket adds 9; adding 25 in the second bracket adds 9\times25=225. Add the same to the right.
(x+3)^2+9(y-5)^2=-225+9+225=9
Divide by 9, then put X=x+3, Y=y-5.
\dfrac{(x+3)^2}{9}+\dfrac{(y-5)^2}{1}=1\;\Longrightarrow\;\dfrac{X^2}{9}+\dfrac{Y^2}{1}=1
The larger denominator is under X^2, so the major axis is horizontal. Compare with \dfrac{X^2}{a^2}+\dfrac{Y^2}{b^2}=1.
a^2=9,\ \ b^2=1\;\Longrightarrow\;a=3,\ \ b=1
Find c from c^2=a^2-b^2 (the distance from the centre to each focus).
c=\sqrt{9-1}=2 \sqrt{2}
Centre. It is where X=0 and Y=0.
C\left(-3,\ 5\right)
Vertices are a units left and right of the centre, (h\pm a,\,k).
A_1\left(-6,\ 5\right),\qquad A_2\left(0,\ 5\right)
Co-vertices are b units below and above the centre, (h,\,k\pm b).
B_1\left(-3,\ 4\right),\qquad B_2\left(-3,\ 6\right)
Foci are on the major axis, c units from the centre, (h\pm c,\,k).
F_1\left(-3 - 2 \sqrt{2},\ 5\right),\qquad F_2\left(-3 + 2 \sqrt{2},\ 5\right)
Eccentricity.
e=\dfrac ca=\dfrac{2 \sqrt{2}}{3}=\dfrac{2 \sqrt{2}}{3}\approx0.943\ (<1)
Major axis: length 2a; it is the line through the foci.
2a=6,\qquad\text{equation } y=5
Minor axis: length 2b; it is the line through the centre perpendicular to the major axis.
2b=2,\qquad\text{equation } x=-3
Directrices are perpendicular to the major axis at distance \dfrac ae=\dfrac{a^2}{c} from the centre.
\dfrac{a^2}{c}=\dfrac{9}{2 \sqrt{2}}=\dfrac{9 \sqrt{2}}{4}\;\Longrightarrow\;x=-3\pm\dfrac{9 \sqrt{2}}{4}
Length of each latus rectum.
\dfrac{2b^2}{a}=\dfrac{2\left(1\right)}{3}=\dfrac{2}{3}
Graph. Plot the centre, the two vertices and the two co-vertices, and join the four points with a smooth oval. Mark the foci on the major axis and draw the directrices as dashed lines.

Dotted: major axis. Dashed: directrices x=-3\pm\tfrac{9 \sqrt{2}}{4}. F_1, F_2 are the foci.

\begin{aligned}&C\left(-3,\ 5\right)\\ &A\left(-6,\ 5\right),\,A'\left(0,\ 5\right)\\ &B\left(-3,\ 4\right),\,B'\left(-3,\ 6\right)\\ &F\left(-3 - 2 \sqrt{2},\ 5\right),\,F'\left(-3 + 2 \sqrt{2},\ 5\right)\\ &e=\tfrac{2 \sqrt{2}}{3}\\ &\text{major axis } 2a=6,\ y=5\\ &\text{minor axis } 2b=2,\ x=-3\\ &\text{directrices } x=-3\pm\tfrac{9 \sqrt{2}}{4}\\ &\text{latus rectum}=\tfrac{2}{3}\end{aligned}

(v) 16x^2+9y^2-32x+36y-92=0

Group the terms and take out the coefficients of x^2 and y^2.
16\left(x^2-2x\right)+9\left(y^2+4y\right)=92
Complete both squares. Adding 1 in the first bracket adds 16; adding 4 in the second adds 36.
16(x-1)^2+9(y+2)^2=92+16+36=144
Divide by 144, then put X=x-1, Y=y+2.
\dfrac{(x-1)^2}{9}+\dfrac{(y+2)^2}{16}=1\;\Longrightarrow\;\dfrac{X^2}{9}+\dfrac{Y^2}{16}=1
The larger denominator is under Y^2, so the major axis is vertical. Compare with \dfrac{X^2}{b^2}+\dfrac{Y^2}{a^2}=1.
a^2=16,\ \ b^2=9\;\Longrightarrow\;a=4,\ \ b=3
Find c from c^2=a^2-b^2 (the distance from the centre to each focus).
c=\sqrt{16-9}=\sqrt{7}
Centre. It is where X=0 and Y=0.
C\left(1,\ -2\right)
Vertices are a units below and above the centre, (h,\,k\pm a).
A_1\left(1,\ -6\right),\qquad A_2\left(1,\ 2\right)
Co-vertices are b units left and right of the centre, (h\pm b,\,k).
B_1\left(-2,\ -2\right),\qquad B_2\left(4,\ -2\right)
Foci are on the major axis, c units from the centre, (h,\,k\pm c).
F_1\left(1,\ - \sqrt{7} - 2\right),\qquad F_2\left(1,\ -2 + \sqrt{7}\right)
Eccentricity.
e=\dfrac ca=\dfrac{\sqrt{7}}{4}=\dfrac{\sqrt{7}}{4}\approx0.661\ (<1)
Major axis: length 2a; it is the line through the foci.
2a=8,\qquad\text{equation } x=1
Minor axis: length 2b; it is the line through the centre perpendicular to the major axis.
2b=6,\qquad\text{equation } y=-2
Directrices are perpendicular to the major axis at distance \dfrac ae=\dfrac{a^2}{c} from the centre.
\dfrac{a^2}{c}=\dfrac{16}{\sqrt{7}}=\dfrac{16 \sqrt{7}}{7}\;\Longrightarrow\;y=-2\pm\dfrac{16 \sqrt{7}}{7}
Length of each latus rectum.
\dfrac{2b^2}{a}=\dfrac{2\left(9\right)}{4}=\dfrac{9}{2}
Graph. Plot the centre, the two vertices and the two co-vertices, and join the four points with a smooth oval. Mark the foci on the major axis and draw the directrices as dashed lines.

Dotted: major axis. Dashed: directrices y=-2\pm\tfrac{16 \sqrt{7}}{7}. F_1, F_2 are the foci.

\begin{aligned}&C\left(1,\ -2\right)\\ &A\left(1,\ -6\right),\,A'\left(1,\ 2\right)\\ &B\left(-2,\ -2\right),\,B'\left(4,\ -2\right)\\ &F\left(1,\ - \sqrt{7} - 2\right),\,F'\left(1,\ -2 + \sqrt{7}\right)\\ &e=\tfrac{\sqrt{7}}{4}\\ &\text{major axis } 2a=8,\ x=1\\ &\text{minor axis } 2b=6,\ y=-2\\ &\text{directrices } y=-2\pm\tfrac{16 \sqrt{7}}{7}\\ &\text{latus rectum}=\tfrac{9}{2}\end{aligned}

From the given information, find the equation of ellipse in each of the following.

Solution

Method. First decide the direction of the major axis from the data, then find the centre (h,k), a and b. Useful relations: c^2=a^2-b^2, e=\dfrac ca so b^2=a^2\left(1-e^2\right), directrix at distance \dfrac ae from the centre, latus rectum \dfrac{2b^2}{a}. When only one focus and its directrix are given, use |PF|=e\,|PM| directly.

(i) Vertices (-4,6), (-16,6) and co-vertices (-10,2), (-10,10)

The vertices have the same y-coordinate, so the major axis is horizontal. The centre is the midpoint of the vertices.
C=\left(\dfrac{-4-16}{2},\ \dfrac{6+6}{2}\right)=(-10,\,6)
a is half the distance between the vertices.
2a=|-4-(-16)|=12\;\Longrightarrow\;a=6,\ a^2=36
b is half the distance between the co-vertices (their midpoint is also (-10,6), as it should be).
2b=|10-2|=8\;\Longrightarrow\;b=4,\ b^2=16
Major axis horizontal, so use \dfrac{(x-h)^2}{a^2}+\dfrac{(y-k)^2}{b^2}=1.
\dfrac{(x+10)^2}{36}+\dfrac{(y-6)^2}{16}=1
Multiply by 144 and expand for the general form.
4(x+10)^2+9(y-6)^2=144\;\Longrightarrow\;4x^2+9y^2+80x-108y+580=0

\dfrac{(x+10)^2}{36}+\dfrac{(y-6)^2}{16}=1\quad\text{or}\quad4x^2+9y^2+80x-108y+580=0

(ii) Centre at (-8,5); vertex (-8,15); focus \left(-8,5+\sqrt5\right)

The centre, vertex and focus all have x=-8, so the major axis is the vertical line x=-8.
a is the distance from the centre to the vertex, and c the distance from the centre to the focus.
a=15-5=10,\qquad c=\left(5+\sqrt5\right)-5=\sqrt5
Find b^2 from b^2=a^2-c^2.
b^2=100-5=95
Major axis vertical, so use \dfrac{(x-h)^2}{b^2}+\dfrac{(y-k)^2}{a^2}=1.
\dfrac{(x+8)^2}{95}+\dfrac{(y-5)^2}{100}=1
Multiply by 1900 and expand.
20(x+8)^2+19(y-5)^2=1900\;\Longrightarrow\;20x^2+19y^2+320x-190y-145=0

The foci \left(-8,5\pm\sqrt5\right) are close to the centre, so the ellipse is nearly a circle (e=\tfrac{\sqrt5}{10}\approx0.22).

\dfrac{(x+8)^2}{95}+\dfrac{(y-5)^2}{100}=1\quad\text{or}\quad20x^2+19y^2+320x-190y-145=0

(iii) Eccentricity is \dfrac{\sqrt{15}}{4}; centre (-5,5); co-vertex (-8,5)

The co-vertex is on the horizontal line y=5 through the centre, so the minor axis is horizontal and the major axis is vertical (x=-5).
b is the distance from the centre to the co-vertex.
b=|-5-(-8)|=3,\qquad b^2=9
Use b^2=a^2-c^2=a^2-a^2e^2=a^2\left(1-e^2\right).
9=a^2\left(1-\dfrac{15}{16}\right)=\dfrac{a^2}{16}\;\Longrightarrow\;a^2=144,\ a=12
Major axis vertical.
\dfrac{(x+5)^2}{9}+\dfrac{(y-5)^2}{144}=1
Multiply by 144 and expand.
16(x+5)^2+(y-5)^2=144\;\Longrightarrow\;16x^2+y^2+160x-10y+281=0

A long, thin ellipse: e=\tfrac{\sqrt{15}}4\approx0.97 is close to 1.

\dfrac{(x+5)^2}{9}+\dfrac{(y-5)^2}{144}=1\quad\text{or}\quad16x^2+y^2+160x-10y+281=0

(iv) Eccentricity is \dfrac34 and passes through the point (3,1) with centre at (0,0); major axis along y-axis.

Centre at the origin and major axis along the y-axis, so the equation has the form
\dfrac{x^2}{b^2}+\dfrac{y^2}{a^2}=1
Relate b^2 to a^2 using the eccentricity.
b^2=a^2\left(1-e^2\right)=a^2\left(1-\dfrac9{16}\right)=\dfrac{7a^2}{16}
So the equation becomes
\dfrac{16x^2}{7a^2}+\dfrac{y^2}{a^2}=1\;\Longrightarrow\;16x^2+7y^2=7a^2
The point (3,1) is on the ellipse. Substitute it.
16(9)+7(1)=7a^2\;\Longrightarrow\;7a^2=151\;\Longrightarrow\;a^2=\dfrac{151}{7},\quad b^2=\dfrac{7}{16}\cdot\dfrac{151}{7}=\dfrac{151}{16}
Write the equation.
\dfrac{x^2}{151/16}+\dfrac{y^2}{151/7}=1\;\Longrightarrow\;16x^2+7y^2=151

a=\sqrt{\tfrac{151}{7}}\approx4.64, b=\tfrac{\sqrt{151}}{4}\approx3.07; the point (3,1) lies on the curve.

\dfrac{16x^2}{151}+\dfrac{7y^2}{151}=1\quad\text{or}\quad16x^2+7y^2=151

(v) Focus at (3,4); directrix x=5 and eccentricity \dfrac23.

Only one focus and its directrix are given, so use the focus-directrix property of a conic: for every point P(x,y) on the ellipse, |PF|=e\,|PM|, where |PM| is the distance from P to the directrix.
Here |PF|=\sqrt{(x-3)^2+(y-4)^2} and |PM|=|x-5|. Square both sides of |PF|=\tfrac23|PM|.
(x-3)^2+(y-4)^2=\dfrac49(x-5)^2
Multiply by 9 and expand.
9x^2-54x+81+9y^2-72y+144=4x^2-40x+100
Collect all terms on the left.
5x^2+9y^2-14x-72y+125=0
Standard form (to see the ellipse). Complete the squares.
5\left(x-\dfrac75\right)^2+9(y-4)^2=\dfrac{49}{5}+144-125=\dfrac{144}{5}
Divide by \dfrac{144}{5}.
\dfrac{\left(x-\frac75\right)^2}{144/25}+\dfrac{(y-4)^2}{16/5}=1
Check. a=\tfrac{12}{5}, c=ae=\tfrac{8}{5}, centre \left(\tfrac75,4\right). The focus is \tfrac75+\tfrac85=3, giving (3,4), and the directrix is x=\tfrac75+\tfrac ae=\tfrac75+\tfrac{18}5=5. Both agree with the data.

The given focus F(3,4) and directrix x=5 are on the right. The other focus is F'\left(-\tfrac15,4\right) and the other directrix is x=-\tfrac{11}{5}.

5x^2+9y^2-14x-72y+125=0\quad\text{or}\quad\dfrac{\left(x-\frac75\right)^2}{144/25}+\dfrac{(y-4)^2}{16/5}=1

(vi) Centre at (1,-1); horizontal tangents are y=7 and y=-9 and vertical tangents are x=6 and x=-4.

The horizontal tangents touch the ellipse at the ends of its vertical axis, and the vertical tangents touch it at the ends of its horizontal axis.
Vertical semi-axis: half the distance between y=7 and y=-9 (each is 8 from the centre y=-1).
\dfrac{7-(-9)}{2}=8
Horizontal semi-axis: half the distance between x=6 and x=-4 (each is 5 from x=1).
\dfrac{6-(-4)}{2}=5
8>5, so the major axis is vertical: a=8, b=5.
\dfrac{(x-1)^2}{25}+\dfrac{(y+1)^2}{64}=1
Multiply by 1600 and expand.
64(x-1)^2+25(y+1)^2=1600\;\Longrightarrow\;64x^2+25y^2-128x+50y-1511=0

The four given tangents form a rectangle around the ellipse; they touch it at the vertices and co-vertices.

\dfrac{(x-1)^2}{25}+\dfrac{(y+1)^2}{64}=1\quad\text{or}\quad64x^2+25y^2-128x+50y-1511=0

(vii) Length of latus rectum is 4; centre at (0,0) and eccentricity is \dfrac13; major axis along y-axis.

Centre at the origin, major axis along the y-axis.
\dfrac{x^2}{b^2}+\dfrac{y^2}{a^2}=1
From the eccentricity:
b^2=a^2\left(1-e^2\right)=a^2\left(1-\dfrac19\right)=\dfrac{8a^2}{9}
Use the length of the latus rectum, \dfrac{2b^2}{a}=4.
\dfrac{2}{a}\cdot\dfrac{8a^2}{9}=4\;\Longrightarrow\;\dfrac{16a}{9}=4\;\Longrightarrow\;a=\dfrac94
Then
a^2=\dfrac{81}{16},\qquad b^2=\dfrac89\cdot\dfrac{81}{16}=\dfrac92
Write the equation and clear fractions (multiply by 81).
\dfrac{x^2}{9/2}+\dfrac{y^2}{81/16}=1\;\Longrightarrow\;\dfrac{2x^2}{9}+\dfrac{16y^2}{81}=1\;\Longrightarrow\;18x^2+16y^2=81
Check. c=ae=\tfrac34; the latus rectum through \left(0,\tfrac34\right) has ends \left(\pm\tfrac{b^2}{a},\tfrac34\right)=\left(\pm2,\tfrac34\right), so its length is 4.

The latus rectum through F_2\left(0,\tfrac34\right) runs from x=-2 to x=2: length 4. Dashed: directrices y=\pm\tfrac{27}{4}.

\dfrac{x^2}{9/2}+\dfrac{y^2}{81/16}=1\quad\text{or}\quad18x^2+16y^2=81

(viii) Centre at (1,1); eccentricity is \dfrac25 and one of the directrices is x=5.

A directrix is perpendicular to the major axis. The directrix x=5 is vertical, so the major axis is horizontal (the line y=1).
\dfrac{(x-1)^2}{a^2}+\dfrac{(y-1)^2}{b^2}=1
The directrices are at distance \dfrac ae from the centre. The centre has x=1, so
\dfrac ae=5-1=4\;\Longrightarrow\;a=4e=4\cdot\dfrac25=\dfrac85,\qquad a^2=\dfrac{64}{25}
Find b^2.
b^2=a^2\left(1-e^2\right)=\dfrac{64}{25}\left(1-\dfrac4{25}\right)=\dfrac{64}{25}\cdot\dfrac{21}{25}=\dfrac{1344}{625}
Write the equation.
\dfrac{(x-1)^2}{64/25}+\dfrac{(y-1)^2}{1344/625}=1
Multiply by 1344 and expand for the general form.
525(x-1)^2+625(y-1)^2=1344\;\Longrightarrow\;525x^2+625y^2-1050x-1250y-194=0

a=\tfrac85, b=\tfrac{8\sqrt{21}}{25}\approx1.47, foci \left(1\pm\tfrac{16}{25},1\right). Dashed: directrices x=5 (given) and x=-3.

\dfrac{(x-1)^2}{64/25}+\dfrac{(y-1)^2}{1344/625}=1\quad\text{or}\quad525x^2+625y^2-1050x-1250y-194=0

Note. In (v) the ellipse is found from the focus-directrix definition; the given focus and directrix are the right-hand pair. In (vi) 'horizontal tangents' and 'vertical tangents' mean the tangent lines parallel to the axes, which touch the ellipse at its vertices and co-vertices.

Find the coordinates of the vertices and co-vertices of the following ellipse by differentiating both sides of equations and solving for horizontal and vertical tangent lines

Solution

Idea. At the four ends of the axes of an ellipse, the tangent is parallel to a coordinate axis. Where \dfrac{dy}{dx}=0 the tangent is horizontal (top and bottom points); where \dfrac{dy}{dx} does not exist (denominator zero) the tangent is vertical (left and right points). The pair that is farther apart gives the vertices; the other pair gives the co-vertices.

(i) 2x^2+3y^2+2x-3y-5=0

Differentiate both sides with respect to x.
4x+6y\dfrac{dy}{dx}+2-3\dfrac{dy}{dx}=0\;\Longrightarrow\;\dfrac{dy}{dx}=-\dfrac{4x+2}{6y-3}
Horizontal tangents: \dfrac{dy}{dx}=0, so the numerator is zero.
4x+2=0\;\Longrightarrow\;x=-\dfrac12
Put x=-\tfrac12 in the equation and multiply by 2.
\dfrac12+3y^2-1-3y-5=0\;\Longrightarrow\;6y^2-6y-11=0\;\Longrightarrow\;y=\dfrac{6\pm\sqrt{36+264}}{12}=\dfrac{3\pm5\sqrt3}{6}
Vertical tangents: \dfrac{dy}{dx} does not exist, so the denominator is zero.
6y-3=0\;\Longrightarrow\;y=\dfrac12
Put y=\tfrac12 in the equation and multiply by 4.
2x^2+\dfrac34+2x-\dfrac32-5=0\;\Longrightarrow\;8x^2+8x-23=0\;\Longrightarrow\;x=\dfrac{-8\pm\sqrt{64+736}}{16}=\dfrac{-2\pm5\sqrt2}{4}
Which are the vertices? The points with vertical tangents are \dfrac{5\sqrt2}{2}\approx3.54 apart (horizontal axis); the points with horizontal tangents are \dfrac{5\sqrt3}{3}\approx2.89 apart (vertical axis). The longer one is the major axis, so it is horizontal.
Vertices (ends of the major axis) and co-vertices:
\text{Vertices } \left(\dfrac{-2\pm5\sqrt2}{4},\ \dfrac12\right),\qquad\text{Co-vertices } \left(-\dfrac12,\ \dfrac{3\pm5\sqrt3}{6}\right)
Check. Completing the squares gives \dfrac{\left(x+\frac12\right)^2}{25/8}+\dfrac{\left(y-\frac12\right)^2}{25/12}=1, so a=\dfrac{5}{2\sqrt2}=\dfrac{5\sqrt2}{4} (horizontal) and b=\dfrac{5}{2\sqrt3}=\dfrac{5\sqrt3}{6}, which agrees.

Dashed: the horizontal tangents y=\tfrac{3\pm5\sqrt3}{6} and the vertical tangents x=\tfrac{-2\pm5\sqrt2}{4}. They touch at the vertices A_1, A_2 and the co-vertices B_1, B_2.

(ii) \dfrac{(x-1)^2}{9}+\dfrac{(y+2)^2}{16}=1

Differentiate both sides with respect to x.
\dfrac{2(x-1)}{9}+\dfrac{2(y+2)}{16}\dfrac{dy}{dx}=0\;\Longrightarrow\;\dfrac{dy}{dx}=-\dfrac{16(x-1)}{9(y+2)}
Horizontal tangents: numerator zero.
x-1=0\;\Longrightarrow\;x=1
Put x=1 in the equation.
\dfrac{(y+2)^2}{16}=1\;\Longrightarrow\;y+2=\pm4\;\Longrightarrow\;y=2\ \text{or}\ -6
Vertical tangents: denominator zero.
y+2=0\;\Longrightarrow\;y=-2
Put y=-2 in the equation.
\dfrac{(x-1)^2}{9}=1\;\Longrightarrow\;x-1=\pm3\;\Longrightarrow\;x=4\ \text{or}\ -2
The points (1,2) and (1,-6) are 8 apart; the points (4,-2) and (-2,-2) are 6 apart. The longer (vertical) axis is the major axis.
So
\text{Vertices } (1,2),\ (1,-6);\qquad\text{Co-vertices } (4,-2),\ (-2,-2)

Horizontal tangents y=2, y=-6 touch at the vertices; vertical tangents x=4, x=-2 touch at the co-vertices.

(iii) \dfrac12x^2+\dfrac34y^2+x-y-5=0

Differentiate both sides with respect to x.
x+\dfrac32y\dfrac{dy}{dx}+1-\dfrac{dy}{dx}=0\;\Longrightarrow\;\dfrac{dy}{dx}=-\dfrac{x+1}{\frac32y-1}=-\dfrac{2(x+1)}{3y-2}
Horizontal tangents: numerator zero.
x+1=0\;\Longrightarrow\;x=-1
Put x=-1 in the equation and multiply by 4.
\dfrac12+\dfrac34y^2-1-y-5=0\;\Longrightarrow\;3y^2-4y-22=0\;\Longrightarrow\;y=\dfrac{4\pm\sqrt{16+264}}{6}=\dfrac{2\pm\sqrt{70}}{3}
Vertical tangents: denominator zero.
3y-2=0\;\Longrightarrow\;y=\dfrac23
Put y=\tfrac23 in the equation and multiply by 6.
\dfrac12x^2+\dfrac13+x-\dfrac23-5=0\;\Longrightarrow\;3x^2+6x-32=0\;\Longrightarrow\;x=\dfrac{-6\pm\sqrt{36+384}}{6}=\dfrac{-3\pm\sqrt{105}}{3}
The vertical-tangent points are \dfrac{2\sqrt{105}}{3}\approx6.83 apart; the horizontal-tangent points are \dfrac{2\sqrt{70}}{3}\approx5.58 apart. So the major axis is horizontal.
So
\text{Vertices } \left(\dfrac{-3\pm\sqrt{105}}{3},\ \dfrac23\right),\qquad\text{Co-vertices } \left(-1,\ \dfrac{2\pm\sqrt{70}}{3}\right)
Check. Completing the squares: \dfrac12(x+1)^2+\dfrac34\left(y-\dfrac23\right)^2=\dfrac{35}{6}, so a^2=\dfrac{35}{3}, b^2=\dfrac{70}{9}, and a=\dfrac{\sqrt{105}}{3}, b=\dfrac{\sqrt{70}}{3}, which agrees.

Vertices \approx(-4.42,0.67) and (2.42,0.67); co-vertices \approx(-1,3.46) and (-1,-2.12).

(iv) x^2+2y^2-6x-4=0

Differentiate both sides with respect to x.
2x+4y\dfrac{dy}{dx}-6=0\;\Longrightarrow\;\dfrac{dy}{dx}=\dfrac{6-2x}{4y}=\dfrac{3-x}{2y}
Horizontal tangents: numerator zero.
3-x=0\;\Longrightarrow\;x=3
Put x=3 in the equation.
9+2y^2-18-4=0\;\Longrightarrow\;y^2=\dfrac{13}{2}\;\Longrightarrow\;y=\pm\sqrt{\dfrac{13}{2}}=\pm\dfrac{\sqrt{26}}{2}
Vertical tangents: denominator zero.
2y=0\;\Longrightarrow\;y=0
Put y=0 in the equation.
x^2-6x-4=0\;\Longrightarrow\;x=\dfrac{6\pm\sqrt{36+16}}{2}=3\pm\sqrt{13}
The vertical-tangent points are 2\sqrt{13}\approx7.21 apart; the horizontal-tangent points are \sqrt{26}\approx5.10 apart. So the major axis is horizontal.
So
\text{Vertices } \left(3\pm\sqrt{13},\ 0\right),\qquad\text{Co-vertices } \left(3,\ \pm\dfrac{\sqrt{26}}{2}\right)
Check. (x-3)^2+2y^2=13, that is \dfrac{(x-3)^2}{13}+\dfrac{y^2}{13/2}=1, so a=\sqrt{13} and b=\sqrt{\tfrac{13}{2}}, which agrees.

Vertices \approx(-0.61,0) and (6.61,0); co-vertices \approx(3,\pm2.55).

Exercise 7.7p. 212

Tangent and normal to an ellipse (Sections 7.10 and 7.11). To test a line against an ellipse, substitute the line in the ellipse and look at the discriminant of the quadratic: \text{Disc}>0 two points, \text{Disc}=0 tangent, \text{Disc}<0 no common point. For the tangent at a given point, find \frac{dy}{dx} by implicit differentiation; the normal has slope -\frac{1}{m}.

Find the points of intersection of the line and ellipse and also find the length of chord intercepted: x-y+1=0; x^{2}+2y^{2}+3x-7y-11=0.

Solution

Make y the subject of the line so that it can be substituted into the ellipse.
x-y+1=0\;\Rightarrow\;y=x+1
The common points satisfy both equations, so put y=x+1 in the equation of the ellipse.
x^{2}+2(x+1)^{2}+3x-7(x+1)-11=0
Expand and collect like terms. The x terms cancel because 4x+3x-7x=0.
\begin{aligned}x^{2}+2x^{2}+4x+2+3x-7x-7-11&=0\\3x^{2}-16&=0\end{aligned}
Solve for x.
x^{2}=\dfrac{16}{3}\;\Rightarrow\;x=\pm\dfrac{4}{\sqrt3}=\pm\dfrac{4\sqrt3}{3}
Find y from y=x+1 for each value of x. This gives the two points of intersection A and B.
A\left(\dfrac{4\sqrt3}{3},\,\dfrac{3+4\sqrt3}{3}\right),\qquad B\left(-\dfrac{4\sqrt3}{3},\,\dfrac{3-4\sqrt3}{3}\right)
The chord intercepted is the segment AB. Use the distance formula; both differences equal \frac{8}{\sqrt3} because the line has slope 1.
|AB|=\sqrt{\left(\dfrac{8}{\sqrt3}\right)^{2}+\left(\dfrac{8}{\sqrt3}\right)^{2}}=\sqrt{\dfrac{128}{3}}=\dfrac{8\sqrt2}{\sqrt3}=\dfrac{8\sqrt6}{3}\approx6.53

The line cuts the ellipse at A(2.31,\,3.31) and B(-2.31,\,-1.31); the chord AB has length \frac{8\sqrt6}{3}\approx6.53.

A\left(\tfrac{4\sqrt3}{3},\tfrac{3+4\sqrt3}{3}\right),\ B\left(-\tfrac{4\sqrt3}{3},\tfrac{3-4\sqrt3}{3}\right),\quad |AB|=\dfrac{8\sqrt6}{3}\approx6.53

Find the value of m so that the line 5x+y-20=0,\ m\neq0 touches the ellipse \dfrac{x^{2}}{4}+\dfrac{y^{2}}{5}=1. Also find the point where it touches the ellipse.

Solution

Read the line as 5x+my-20=0 (see the note). Since m\neq0, we can make y the subject.
y=\dfrac{20-5x}{m}
Clear the fractions in the ellipse by multiplying by 20.
\dfrac{x^{2}}{4}+\dfrac{y^{2}}{5}=1\;\Rightarrow\;5x^{2}+4y^{2}=20
Substitute y and multiply through by m^{2}.
5m^{2}x^{2}+4(20-5x)^{2}=20m^{2}
Expand, collect the powers of x and divide by 5.
\begin{aligned}5m^{2}x^{2}+1600-800x+100x^{2}-20m^{2}&=0\\(m^{2}+20)x^{2}-160x+(320-4m^{2})&=0\end{aligned}
The line touches the ellipse when this quadratic has equal roots, that is when its discriminant is zero. Work out the discriminant.
\begin{aligned}\text{Disc}&=(-160)^{2}-4(m^{2}+20)(320-4m^{2})\\&=25600-4\left(240m^{2}-4m^{4}+6400\right)\\&=16m^{4}-960m^{2}=16m^{2}(m^{2}-60)\end{aligned}
Put \text{Disc}=0. The factor m^{2} is not zero because m\neq0.
m^{2}-60=0\;\Rightarrow\;m=\pm\sqrt{60}=\pm2\sqrt{15}
At a tangent the two roots are equal, x=-\dfrac{B}{2A} for the quadratic Ax^{2}+Bx+C=0. Use m^{2}=60.
x=\dfrac{160}{2(m^{2}+20)}=\dfrac{160}{2(80)}=1
Find y from the line.
y=\dfrac{20-5(1)}{m}=\dfrac{15}{m}=\dfrac{15}{\pm2\sqrt{15}}=\pm\dfrac{\sqrt{15}}{2}
Check that the point is on the ellipse: \dfrac14+\dfrac{15/4}{5}=\dfrac14+\dfrac34=1.

For m=\pm2\sqrt{15} the lines touch the ellipse at \left(1,\pm\frac{\sqrt{15}}{2}\right); both pass through (4,0).

m=2\sqrt{15}:\ \left(1,\tfrac{\sqrt{15}}{2}\right);\qquad m=-2\sqrt{15}:\ \left(1,-\tfrac{\sqrt{15}}{2}\right)

Note. Misprint: the printed line 5x+y-20=0 contains no m, and this line does not even meet the ellipse. The condition m\neq0 shows that m is the coefficient of y, so the line is read as 5x+my-20=0.

Find the equation of tangent to the ellipse x^{2}+2y^{2}-3x+5y-3=0 with slope 1.

Solution

A line with slope 1 has the form y=x+c. We must find c.
y=x+c
Substitute y=x+c in the equation of the ellipse.
x^{2}+2(x+c)^{2}-3x+5(x+c)-3=0
Expand and collect the powers of x.
\begin{aligned}x^{2}+2x^{2}+4cx+2c^{2}-3x+5x+5c-3&=0\\3x^{2}+(4c+2)x+(2c^{2}+5c-3)&=0\end{aligned}
For a tangent this quadratic must have equal roots, so its discriminant is zero.
\begin{aligned}(4c+2)^{2}-4(3)(2c^{2}+5c-3)&=0\\16c^{2}+16c+4-24c^{2}-60c+36&=0\\-8c^{2}-44c+40&=0\end{aligned}
Divide by -4 and solve for c with the quadratic formula.
2c^{2}+11c-10=0\;\Rightarrow\;c=\dfrac{-11\pm\sqrt{121+80}}{4}=\dfrac{-11\pm\sqrt{201}}{4}
Put each value in y=x+c and multiply by 4 to clear the fraction.
y=x+\dfrac{-11\pm\sqrt{201}}{4}\;\Rightarrow\;4x-4y-11\pm\sqrt{201}=0

The two tangents of slope 1 touch the ellipse at about (-0.86,\,-0.07) and (3.86,\,-2.43).

4x-4y-11+\sqrt{201}=0\quad\text{and}\quad4x-4y-11-\sqrt{201}=0

Find the equations of tangent to the ellipse 3x^{2}+4y^{2}=12 which are perpendicular to the line y+2x=4.

Solution

Divide by 12 to write the ellipse in standard form.
\dfrac{x^{2}}{4}+\dfrac{y^{2}}{3}=1,\qquad a^{2}=4,\ b^{2}=3
The given line is y=-2x+4 with slope -2. Perpendicular lines have slopes whose product is -1.
m\times(-2)=-1\;\Rightarrow\;m=\dfrac12
Let the tangent be y=\frac12x+c and substitute it in 3x^{2}+4y^{2}=12.
3x^{2}+4\left(\dfrac{x}{2}+c\right)^{2}=12\;\Rightarrow\;4x^{2}+4cx+4c^{2}-12=0\;\Rightarrow\;x^{2}+cx+c^{2}-3=0
A tangent meets the ellipse at one point only, so the discriminant is zero.
c^{2}-4(c^{2}-3)=0\;\Rightarrow\;12-3c^{2}=0\;\Rightarrow\;c=\pm2
Write the two tangents.
y=\dfrac{x}{2}+2\;\Rightarrow\;x-2y+4=0,\qquad y=\dfrac{x}{2}-2\;\Rightarrow\;x-2y-4=0
Points of contact (double root x=-\frac c2): c=2 gives x=-1, y=\frac32; c=-2 gives x=1, y=-\frac32. Both satisfy 3x^{2}+4y^{2}=3+9=12.
\left(-1,\tfrac32\right),\qquad\left(1,-\tfrac32\right)

Both tangents have slope \frac12, so they are perpendicular to the dashed line y+2x=4 (slope -2).

x-2y+4=0\quad\text{and}\quad x-2y-4=0

Find the equations of the tangent and normal to the ellipse 2x^{2}+3y^{2}-5x-10y+5=0 at the point (3,2).

Solution

Check that (3,2) lies on the ellipse.
2(9)+3(4)-5(3)-10(2)+5=18+12-15-20+5=0
Differentiate the equation implicitly with respect to x (the book's method in Example 22).
4x+6y\dfrac{dy}{dx}-5-10\dfrac{dy}{dx}=0
Collect the \frac{dy}{dx} terms and solve.
(6y-10)\dfrac{dy}{dx}=5-4x\;\Rightarrow\;\dfrac{dy}{dx}=\dfrac{5-4x}{6y-10}
At (3,2) this is the slope of the tangent.
m=\dfrac{5-12}{12-10}=-\dfrac72
Point-slope form y-y_{1}=m(x-x_{1}) gives the tangent.
y-2=-\dfrac72(x-3)\;\Rightarrow\;2y-4=-7x+21\;\Rightarrow\;7x+2y-25=0
The normal is perpendicular to the tangent, so its slope is -\frac1m=\frac27.
y-2=\dfrac27(x-3)\;\Rightarrow\;7y-14=2x-6\;\Rightarrow\;2x-7y+8=0

In standard form the ellipse is \dfrac{(x-\frac54)^{2}}{155/48}+\dfrac{(y-\frac53)^{2}}{155/72}=1. The tangent and the normal at (3,2) are perpendicular.

\text{Tangent: }7x+2y-25=0,\qquad\text{Normal: }2x-7y+8=0

Find the equations of the tangents and normal to the ellipse 15x^{2}+6y^{2}-8x-5y+2=0 at the point with ordinate \frac13.

Solution

The ordinate is the y-coordinate. Put y=\frac13 in the equation to find the abscissa (x-coordinate).
15x^{2}+6\left(\tfrac19\right)-8x-5\left(\tfrac13\right)+2=0\;\Rightarrow\;15x^{2}-8x+1=0
Factorise.
(3x-1)(5x-1)=0\;\Rightarrow\;x=\tfrac13\ \text{ or }\ x=\tfrac15
So two points of the ellipse have ordinate \frac13, which is why the question asks for tangents (plural).
P\left(\tfrac13,\tfrac13\right),\qquad Q\left(\tfrac15,\tfrac13\right)
Differentiate the equation implicitly with respect to x.
30x+12y\dfrac{dy}{dx}-8-5\dfrac{dy}{dx}=0\;\Rightarrow\;\dfrac{dy}{dx}=\dfrac{8-30x}{12y-5}
At both points y=\frac13, so 12y-5=-1 and the slope is 30x-8.
m_{P}=30\left(\tfrac13\right)-8=2,\qquad m_{Q}=30\left(\tfrac15\right)-8=-2
Tangent at P (slope 2).
y-\tfrac13=2\left(x-\tfrac13\right)\;\Rightarrow\;3y-1=6x-2\;\Rightarrow\;6x-3y-1=0
Normal at P (slope -\frac12).
y-\tfrac13=-\tfrac12\left(x-\tfrac13\right)\;\Rightarrow\;6y-2=-3x+1\;\Rightarrow\;x+2y-1=0
Tangent at Q (slope -2). Multiply by 15 to clear the fractions.
y-\tfrac13=-2\left(x-\tfrac15\right)\;\Rightarrow\;15y-5=-30x+6\;\Rightarrow\;30x+15y-11=0
Normal at Q (slope \frac12). Multiply by 30.
y-\tfrac13=\tfrac12\left(x-\tfrac15\right)\;\Rightarrow\;30y-10=15x-3\;\Rightarrow\;15x-30y+7=0

Completing the squares gives 15\left(x-\frac{4}{15}\right)^{2}+6\left(y-\frac{5}{12}\right)^{2}=\frac{13}{120}, a very small ellipse. Solid lines: tangents; dashed lines of the same colour: the normals at P and Q.

\begin{aligned}&\text{At }P\left(\tfrac13,\tfrac13\right):\\ &\quad\text{tangent }6x-3y-1=0\\ &\quad\text{normal }x+2y-1=0\\ &\text{At }Q\left(\tfrac15,\tfrac13\right):\\ &\quad\text{tangent }30x+15y-11=0\\ &\quad\text{normal }15x-30y+7=0\end{aligned}

Prove that tangent at any point to the ellipse make equal angles with the line joining the point with the foci (reflecting property of ellipse).

Solution

Idea. Light or sound leaving one focus is reflected by an elliptical mirror to the other focus. This happens because the tangent at any point P makes equal angles with the focal lines PF_{1} and PF_{2}. We prove it with the angle between two lines: if a line of slope m_{1} is turned anticlockwise through \theta onto a line of slope m_{2}, then \tan\theta=\dfrac{m_{2}-m_{1}}{1+m_{1}m_{2}}.

Take the ellipse in standard form, its foci, and any point P(x_{1},y_{1}) on it that is not a vertex (so y_{1}\neq0).
\dfrac{x^{2}}{a^{2}}+\dfrac{y^{2}}{b^{2}}=1,\qquad F_{1}(-c,0),\ F_{2}(c,0),\qquad c^{2}=a^{2}-b^{2}
P lies on the ellipse. Multiply its equation by a^{2}b^{2}; this relation is used below.
b^{2}x_{1}^{2}+a^{2}y_{1}^{2}=a^{2}b^{2}\qquad(1)
Slope of the tangent at P: differentiate b^{2}x^{2}+a^{2}y^{2}=a^{2}b^{2} implicitly.
2b^{2}x+2a^{2}y\dfrac{dy}{dx}=0\;\Rightarrow\;m=-\dfrac{b^{2}x_{1}}{a^{2}y_{1}}
Slopes of the focal lines PF_{1} and PF_{2}.
m_{1}=\dfrac{y_{1}}{x_{1}+c},\qquad m_{2}=\dfrac{y_{1}}{x_{1}-c}
Let \alpha be the angle from PF_{1} to the tangent, \tan\alpha=\dfrac{m-m_{1}}{1+mm_{1}}. Simplify the numerator using (1).
m-m_{1}=-\dfrac{b^{2}x_{1}(x_{1}+c)+a^{2}y_{1}^{2}}{a^{2}y_{1}(x_{1}+c)}=-\dfrac{a^{2}b^{2}+b^{2}cx_{1}}{a^{2}y_{1}(x_{1}+c)}=-\dfrac{b^{2}(a^{2}+cx_{1})}{a^{2}y_{1}(x_{1}+c)}
Simplify the denominator using a^{2}-b^{2}=c^{2}.
1+mm_{1}=\dfrac{a^{2}(x_{1}+c)-b^{2}x_{1}}{a^{2}(x_{1}+c)}=\dfrac{c^{2}x_{1}+a^{2}c}{a^{2}(x_{1}+c)}=\dfrac{c(a^{2}+cx_{1})}{a^{2}(x_{1}+c)}
Divide. The common factors a^{2}+cx_{1} and a^{2}(x_{1}+c) cancel.
\tan\alpha=-\dfrac{b^{2}}{cy_{1}}
Let \beta be the angle from the tangent to PF_{2}, \tan\beta=\dfrac{m_{2}-m}{1+mm_{2}}. Work in the same way.
\begin{aligned}m_{2}-m&=\dfrac{a^{2}y_{1}^{2}+b^{2}x_{1}(x_{1}-c)}{a^{2}y_{1}(x_{1}-c)}=\dfrac{b^{2}(a^{2}-cx_{1})}{a^{2}y_{1}(x_{1}-c)}\\1+mm_{2}&=\dfrac{a^{2}(x_{1}-c)-b^{2}x_{1}}{a^{2}(x_{1}-c)}=\dfrac{c(cx_{1}-a^{2})}{a^{2}(x_{1}-c)}\end{aligned}
Divide, using a^{2}-cx_{1}=-(cx_{1}-a^{2}).
\tan\beta=-\dfrac{b^{2}}{cy_{1}}
So \tan\alpha=\tan\beta. Both angles lie between 0^\circ and 180^\circ, where equal tangents mean equal angles.
\alpha=\beta

Turning from PF_{1} to the tangent needs the same angle as turning from the tangent to PF_{2}. So the tangent at P makes equal angles with PF_{1} and PF_{2}, and the normal at P bisects \angle F_{1}PF_{2}. A ray from F_{1} is therefore reflected through F_{2}. Hence proved.

Example with a=5, b=3, c=4 and P(3,\,2.4): the tangent makes the same angle (about 43^\circ) with PF_{1} and with PF_{2}, and the normal bisects \angle F_{1}PF_{2}.

\tan\alpha=\tan\beta=-\dfrac{b^{2}}{cy_{1}}\;\Rightarrow\;\alpha=\beta

Exercise 7.8p. 219

Hyperbola (Section 7.12). Complete the squares to reach \dfrac{(x-h)^{2}}{a^{2}}-\dfrac{(y-k)^{2}}{b^{2}}=1 or \dfrac{(y-k)^{2}}{a^{2}}-\dfrac{(x-h)^{2}}{b^{2}}=1, put X=x-h, Y=y-k, read the elements and convert back. For a hyperbola c^{2}=a^{2}+b^{2}, e=\frac ca>1, the directrices are X=\pm\frac ae (or Y=\pm\frac ae) and each latus rectum has length \frac{2b^{2}}{a}.

Find the centre, vertices, co-vertices, foci, eccentricity, length and equation of transverse axis, length and equation of conjugate axis, directrices and length of latus rectums for the given equations of hyperbola. Also sketch the hyperbola in each case.

Solution

(i) \dfrac{(x-5)^{2}}{36}-\dfrac{(y-4)^{2}}{81}=1

The equation is already in standard form. The positive term contains x, so the transverse (focal) axis is horizontal. Put X=x-5, Y=y-4.
\dfrac{X^{2}}{6^{2}}-\dfrac{Y^{2}}{9^{2}}=1,\qquad a=6,\ b=9
For a hyperbola c^{2}=a^{2}+b^{2} (not a^{2}-b^{2} as for an ellipse).
c^{2}=36+81=117\;\Rightarrow\;c=3\sqrt{13}\approx10.82
Eccentricity e=\frac{c}{a} and the distance \frac ae from the centre to each directrix.
e=\dfrac{3\sqrt{13}}{6}=\dfrac{\sqrt{13}}{2}\approx1.80,\qquad\dfrac ae=\dfrac{6}{\sqrt{13}/2}=\dfrac{12}{\sqrt{13}}\approx3.33
Convert each element back with x=X+5, y=Y+4.

Element	Rule in X,Y	Result
Centre	X=0,\ Y=0	(5,4)
Vertices	(\pm a,0)	(-1,4) and (11,4)
Co-vertices	(0,\pm b)	(5,-5) and (5,13)
Foci	(\pm c,0)	(5-3\sqrt{13},\,4) and (5+3\sqrt{13},\,4)
Eccentricity	e=\frac ca	\frac{\sqrt{13}}{2}\approx1.80
Transverse axis	Y=0, length 2a	y=4, length 12
Conjugate axis	X=0, length 2b	x=5, length 18
Directrices	X=\pm\frac ae	x=5\pm\frac{12}{\sqrt{13}} (x\approx8.33, x\approx1.67)
Latus rectum	\frac{2b^{2}}{a}	\frac{2(81)}{6}=27

Sketch. Plot the centre, vertices and co-vertices, draw the rectangle through them and its diagonals (dashed guide lines, called asymptotes), then draw the two branches through the vertices so that they open away from the centre and approach the diagonals.

Branches open left and right. Filled dots off the curve: foci F_{1,2}\approx(-5.82,4),\,(15.82,4); open dots: co-vertices; dotted lines: directrices.

(ii) \dfrac{(y-10)^{2}}{10}-\dfrac{(x-1)^{2}}{15}=1

The positive term contains y, so the transverse axis is vertical. Put X=x-1, Y=y-10.
\dfrac{Y^{2}}{(\sqrt{10})^{2}}-\dfrac{X^{2}}{(\sqrt{15})^{2}}=1,\qquad a=\sqrt{10},\ b=\sqrt{15}
Find c.
c^{2}=a^{2}+b^{2}=10+15=25\;\Rightarrow\;c=5
Eccentricity and the directrix distance \frac ae.
e=\dfrac{5}{\sqrt{10}}=\dfrac{\sqrt{10}}{2}\approx1.58,\qquad\dfrac ae=\dfrac{\sqrt{10}}{\sqrt{10}/2}=2
For a vertical transverse axis the vertices, foci and directrices are on the Y side: vertices (0,\pm a), co-vertices (\pm b,0), foci (0,\pm c), directrices Y=\pm\frac ae. Convert with x=X+1, y=Y+10.

Element	Rule in X,Y	Result
Centre	X=0,\ Y=0	(1,10)
Vertices	(0,\pm a)	(1,10+\sqrt{10}) and (1,10-\sqrt{10})
Co-vertices	(\pm b,0)	(1+\sqrt{15},10) and (1-\sqrt{15},10)
Foci	(0,\pm c)	(1,15) and (1,5)
Eccentricity	e=\frac ca	\frac{\sqrt{10}}{2}\approx1.58
Transverse axis	X=0, length 2a	x=1, length 2\sqrt{10}\approx6.32
Conjugate axis	Y=0, length 2b	y=10, length 2\sqrt{15}\approx7.75
Directrices	Y=\pm\frac ae	y=12 and y=8
Latus rectum	\frac{2b^{2}}{a}	\frac{30}{\sqrt{10}}=3\sqrt{10}\approx9.49

Branches open up and down. Vertices A_{1,2}=(1,10\mp\sqrt{10}), co-vertices B_{1,2}=(1\mp\sqrt{15},10); dotted lines: directrices y=8 and y=12.

(iii) -16x^{2}+9y^{2}+32x+144y-16=0

Group the y terms and the x terms and move the constant to the right.
9(y^{2}+16y)-16(x^{2}-2x)=16
Complete each square: y^{2}+16y=(y+8)^{2}-64 and x^{2}-2x=(x-1)^{2}-1.
9\left[(y+8)^{2}-64\right]-16\left[(x-1)^{2}-1\right]=16
Simplify.
9(y+8)^{2}-576-16(x-1)^{2}+16=16\;\Rightarrow\;9(y+8)^{2}-16(x-1)^{2}=576
Divide by 576 to get the standard form.
\dfrac{(y+8)^{2}}{64}-\dfrac{(x-1)^{2}}{36}=1
The y term is positive, so the transverse axis is vertical. Put X=x-1, Y=y+8.
a=8,\quad b=6,\quad c=\sqrt{64+36}=10,\quad e=\dfrac{10}{8}=\dfrac54,\quad\dfrac ae=\dfrac{8}{5/4}=\dfrac{32}{5}

Element	Rule in X,Y	Result
Centre	X=0,\ Y=0	(1,-8)
Vertices	(0,\pm a)	(1,0) and (1,-16)
Co-vertices	(\pm b,0)	(7,-8) and (-5,-8)
Foci	(0,\pm c)	(1,2) and (1,-18)
Eccentricity	e=\frac ca	\frac54
Transverse axis	X=0, length 2a	x=1, length 16
Conjugate axis	Y=0, length 2b	y=-8, length 12
Directrices	Y=\pm\frac ae	y=-8\pm\frac{32}{5}, that is y=-\frac85 and y=-\frac{72}{5}
Latus rectum	\frac{2b^{2}}{a}	\frac{2(36)}{8}=9

Centre C(1,-8). Branches open up and down; dotted lines: directrices y=-1.6 and y=-14.4.

(iv) x^{2}-4y^{2}+2x-40y-135=0

Group the terms and move the constant to the right.
(x^{2}+2x)-4(y^{2}+10y)=135
Complete each square: x^{2}+2x=(x+1)^{2}-1 and y^{2}+10y=(y+5)^{2}-25.
\left[(x+1)^{2}-1\right]-4\left[(y+5)^{2}-25\right]=135
Simplify.
(x+1)^{2}-1-4(y+5)^{2}+100=135\;\Rightarrow\;(x+1)^{2}-4(y+5)^{2}=36
Divide by 36.
\dfrac{(x+1)^{2}}{36}-\dfrac{(y+5)^{2}}{9}=1
The x term is positive, so the transverse axis is horizontal. Put X=x+1, Y=y+5.
a=6,\quad b=3,\quad c=\sqrt{36+9}=3\sqrt5,\quad e=\dfrac{3\sqrt5}{6}=\dfrac{\sqrt5}{2},\quad\dfrac ae=\dfrac{12}{\sqrt5}

Element	Rule in X,Y	Result
Centre	X=0,\ Y=0	(-1,-5)
Vertices	(\pm a,0)	(5,-5) and (-7,-5)
Co-vertices	(0,\pm b)	(-1,-2) and (-1,-8)
Foci	(\pm c,0)	(-1+3\sqrt5,\,-5) and (-1-3\sqrt5,\,-5)
Eccentricity	e=\frac ca	\frac{\sqrt5}{2}\approx1.12
Transverse axis	Y=0, length 2a	y=-5, length 12
Conjugate axis	X=0, length 2b	x=-1, length 6
Directrices	X=\pm\frac ae	x=-1\pm\frac{12}{\sqrt5} (x\approx4.37, x\approx-6.37)
Latus rectum	\frac{2b^{2}}{a}	\frac{2(9)}{6}=3

Centre (-1,-5); foci F_{1,2}\approx(-7.71,-5),\,(5.71,-5); dotted lines: directrices.

Note. The book's summary table (Section 7.12) and its Example 23 give the directrices of a hyperbola as x=\pm\frac ca. That is a misprint, because \frac ca is the eccentricity. The directrices are X=\pm\frac ae=\pm\frac{a^{2}}{c} (the second form in the same table), and this is used below. For Example 23 they are x=2\pm\frac{9}{\sqrt{13}}.

From the given information find the equation of hyperbola in each of the following.

Solution

(i) Vertices (8,14), (8,-10); conjugate axis of length 6 units.

Both vertices have x=8, so the transverse axis is the vertical line x=8 and the equation has the form:
\dfrac{(y-k)^{2}}{a^{2}}-\dfrac{(x-h)^{2}}{b^{2}}=1
The centre is the midpoint of the vertices.
(h,k)=\left(8,\ \dfrac{14+(-10)}{2}\right)=(8,2)
The distance between the vertices is 2a, and the length of the conjugate axis is 2b.
2a=14-(-10)=24\;\Rightarrow\;a=12,\qquad2b=6\;\Rightarrow\;b=3
Substitute.
\dfrac{(y-2)^{2}}{144}-\dfrac{(x-8)^{2}}{9}=1
General form (multiply by 144 and expand), if required.
(y-2)^{2}-16(x-8)^{2}=144\;\Rightarrow\;16x^{2}-y^{2}-256x+4y+1164=0

A narrow hyperbola: a=12 is much larger than b=3. Open dots: ends of the conjugate axis (5,2) and (11,2).

\dfrac{(y-2)^{2}}{144}-\dfrac{(x-8)^{2}}{9}=1

(ii) Vertices \left(-2,\frac52\right), \left(-16,\frac52\right) and end points of conjugate axis are \left(-9,\frac{15}{2}\right); \left(-9,-\frac52\right).

The vertices have the same y-coordinate, so the transverse axis is horizontal.
\dfrac{(x-h)^{2}}{a^{2}}-\dfrac{(y-k)^{2}}{b^{2}}=1
The centre is the midpoint of the vertices. It is also the midpoint of the ends of the conjugate axis, which confirms the data.
(h,k)=\left(\dfrac{-2+(-16)}{2},\ \dfrac52\right)=\left(-9,\ \dfrac52\right)
a is half the distance between the vertices; b is half the length of the conjugate axis.
a=\dfrac{-2-(-16)}{2}=7,\qquad b=\dfrac{\frac{15}{2}-\left(-\frac52\right)}{2}=5
Substitute.
\dfrac{(x+9)^{2}}{49}-\dfrac{\left(y-\frac52\right)^{2}}{25}=1

The vertices and the ends of the conjugate axis fix the rectangle; the branches pass through the vertices.

\dfrac{(x+9)^{2}}{49}-\dfrac{(y-\frac52)^{2}}{25}=1

(iii) Vertices are (-5,1), (-5,-7) and foci are (-5,-3+\sqrt{97}), (-5,-3-\sqrt{97}).

The vertices and foci all have x=-5, so the transverse axis is vertical.
\dfrac{(y-k)^{2}}{a^{2}}-\dfrac{(x-h)^{2}}{b^{2}}=1
The centre is the midpoint of the vertices.
(h,k)=\left(-5,\ \dfrac{1+(-7)}{2}\right)=(-5,-3)
The distance from the centre to a vertex is a and to a focus is c.
a=1-(-3)=4,\qquad c=\sqrt{97}
For a hyperbola b^{2}=c^{2}-a^{2}.
b^{2}=97-16=81
Substitute.
\dfrac{(y+3)^{2}}{16}-\dfrac{(x+5)^{2}}{81}=1

Foci F_{1,2}=(-5,-3\mp\sqrt{97})\approx(-5,-12.85),\,(-5,6.85).

\dfrac{(y+3)^{2}}{16}-\dfrac{(x+5)^{2}}{81}=1

(iv) Foci are (8,-5+\sqrt{53}), (8,-5-\sqrt{53}) and end points of conjugate axis are (15,5); (1,-5).

Both foci have x=8, so the transverse axis is vertical.
\dfrac{(y-k)^{2}}{a^{2}}-\dfrac{(x-h)^{2}}{b^{2}}=1
The centre is the midpoint of the foci, and the distance from the centre to a focus is c.
(h,k)=(8,-5),\qquad c=\sqrt{53}
The ends of the conjugate axis are (15,-5) and (1,-5); half the distance between them is b.
b=\dfrac{15-1}{2}=7
For a hyperbola a^{2}=c^{2}-b^{2}.
a^{2}=53-49=4\;\Rightarrow\;a=2
Substitute.
\dfrac{(y+5)^{2}}{4}-\dfrac{(x-8)^{2}}{49}=1

A wide hyperbola: b=7 is larger than a=2. Vertices (8,-3) and (8,-7); foci (8,-5\pm\sqrt{53}).

\dfrac{(y+5)^{2}}{4}-\dfrac{(x-8)^{2}}{49}=1

(v) Vertices are (0,9), (0,-9) and passes through the point (8,15).

The vertices are on the y-axis and symmetric about the origin, so the centre is (0,0), the transverse axis is the y-axis and a=9.
\dfrac{y^{2}}{81}-\dfrac{x^{2}}{b^{2}}=1
The point (8,15) lies on the hyperbola, so it satisfies the equation.
\dfrac{225}{81}-\dfrac{64}{b^{2}}=1
Solve for b^{2}.
\dfrac{64}{b^{2}}=\dfrac{225}{81}-1=\dfrac{144}{81}\;\Rightarrow\;b^{2}=\dfrac{64\times81}{144}=36
Substitute. Multiplying by 324 gives a form without fractions.
\dfrac{y^{2}}{81}-\dfrac{x^{2}}{36}=1\quad\text{or}\quad4y^{2}-9x^{2}=324

The branch through (0,9) also passes through (8,15).

\dfrac{y^{2}}{81}-\dfrac{x^{2}}{36}=1

(vi) Eccentricity is \frac54; centre at (1,1) and passes through (-7,2), focal axis is horizontal.

A horizontal focal axis and centre (1,1) give the form:
\dfrac{(x-1)^{2}}{a^{2}}-\dfrac{(y-1)^{2}}{b^{2}}=1\qquad(1)
Link b to a with the eccentricity: c=ae and b^{2}=c^{2}-a^{2}.
b^{2}=a^{2}e^{2}-a^{2}=a^{2}\left(\dfrac{25}{16}-1\right)=\dfrac{9a^{2}}{16}
The point (-7,2) satisfies (1).
\dfrac{(-7-1)^{2}}{a^{2}}-\dfrac{(2-1)^{2}}{b^{2}}=1\;\Rightarrow\;\dfrac{64}{a^{2}}-\dfrac{16}{9a^{2}}=1
Combine the fractions and solve for a^{2}, then find b^{2}.
\dfrac{576-16}{9a^{2}}=1\;\Rightarrow\;a^{2}=\dfrac{560}{9},\qquad b^{2}=\dfrac{9}{16}\cdot\dfrac{560}{9}=35
Substitute in (1).
\dfrac{9(x-1)^{2}}{560}-\dfrac{(y-1)^{2}}{35}=1\quad\text{or}\quad9(x-1)^{2}-16(y-1)^{2}=560
General form, if required.
9x^{2}-16y^{2}-18x+32y-567=0

a=\sqrt{560}/3\approx7.89, b=\sqrt{35}\approx5.92, c=ae\approx9.86. The left branch passes through (-7,2).

\dfrac{9(x-1)^{2}}{560}-\dfrac{(y-1)^{2}}{35}=1

(vii) Focus (5,3); directrix y=-2 and eccentricity \frac53.

Take any point P(x,y) on the hyperbola and let M be the foot of the perpendicular from P to the directrix. By the focus-directrix definition (as in Example 28):
|PF|=e\,|PM|
Write both distances. The directrix y+2=0 is horizontal.
|PF|=\sqrt{(x-5)^{2}+(y-3)^{2}},\qquad|PM|=\dfrac{|y+2|}{\sqrt{0^{2}+1^{2}}}=|y+2|
Substitute e=\frac53, multiply by 3 and square both sides.
3\sqrt{(x-5)^{2}+(y-3)^{2}}=5|y+2|\;\Rightarrow\;9\left[(x-5)^{2}+(y-3)^{2}\right]=25(y+2)^{2}
Expand both sides.
9x^{2}-90x+225+9y^{2}-54y+81=25y^{2}+100y+100
Bring every term to one side.
9x^{2}-16y^{2}-90x-154y+206=0
Check by completing the squares. The transverse axis is vertical (perpendicular to the horizontal directrix).
\dfrac{\left(y+\frac{77}{16}\right)^{2}}{\left(\frac{75}{16}\right)^{2}}-\dfrac{(x-5)^{2}}{\left(\frac{25}{4}\right)^{2}}=1
Here a=\frac{75}{16}, b=\frac{25}{4}, c=\sqrt{a^{2}+b^{2}}=\frac{125}{16}, so e=\frac ca=\frac53. The upper focus is at y=-\frac{77}{16}+\frac{125}{16}=3 and the lower directrix at y=-\frac{77}{16}+\frac{a}{e}=-\frac{77}{16}+\frac{45}{16}=-2, as given.

Centre \left(5,-\frac{77}{16}\right). The given focus F(5,3) belongs to the upper branch, and y=-2 is its directrix.

9x^{2}-16y^{2}-90x-154y+206=0

(viii) Centre at (0,0); length of latus rectum is 5, eccentricity \frac54; conjugate axis along x-axis.

The conjugate axis is the x-axis, so the transverse (focal) axis is the y-axis.
\dfrac{y^{2}}{a^{2}}-\dfrac{x^{2}}{b^{2}}=1
The eccentricity links b and a.
b^{2}=c^{2}-a^{2}=a^{2}(e^{2}-1)=a^{2}\left(\dfrac{25}{16}-1\right)=\dfrac{9a^{2}}{16}
The length of the latus rectum is \frac{2b^{2}}{a}.
\dfrac{2b^{2}}{a}=5\;\Rightarrow\;\dfrac{2}{a}\cdot\dfrac{9a^{2}}{16}=5\;\Rightarrow\;\dfrac{9a}{8}=5\;\Rightarrow\;a=\dfrac{40}{9}
Find a^{2} and b^{2}.
a^{2}=\dfrac{1600}{81},\qquad b^{2}=\dfrac{9}{16}\cdot\dfrac{1600}{81}=\dfrac{100}{9}
Substitute.
\dfrac{81y^{2}}{1600}-\dfrac{9x^{2}}{100}=1\quad\text{or}\quad81y^{2}-144x^{2}=1600

c=ae=\frac{50}{9}. The latus rectum through F_{2}\left(0,\frac{50}{9}\right) runs from x=-2.5 to x=2.5, length 5.

\dfrac{81y^{2}}{1600}-\dfrac{9x^{2}}{100}=1

Note. Part (iv) has a misprint: the end point of the conjugate axis (15,5) should be (15,-5).

Find the vertices of the following hyperbolas by differentiating the given equation and solving for horizontal/vertical tangent lines.

Solution

(i) \dfrac{\left(x-\frac12\right)^{2}}{3}-\dfrac{(y+3)^{2}}{5}=1

Differentiate both sides with respect to x (implicit differentiation).
\dfrac{2\left(x-\frac12\right)}{3}-\dfrac{2(y+3)}{5}\dfrac{dy}{dx}=0\;\Rightarrow\;\dfrac{dy}{dx}=\dfrac{5\left(x-\frac12\right)}{3(y+3)}
Horizontal tangent: \frac{dy}{dx}=0, so the numerator is zero, x=\frac12. Put this in the equation.
-\dfrac{(y+3)^{2}}{5}=1\;\Rightarrow\;(y+3)^{2}=-5
A square cannot be negative, so there is no real point: the curve has no horizontal tangent.
Vertical tangent: the denominator of \frac{dy}{dx} is zero, so y=-3. Put this in the equation.
\dfrac{\left(x-\frac12\right)^{2}}{3}=1\;\Rightarrow\;x-\dfrac12=\pm\sqrt3\;\Rightarrow\;x=\dfrac12\pm\sqrt3
The vertical tangents x=\frac12\pm\sqrt3 touch the hyperbola at the ends of its transverse axis, which are the vertices.
\left(\tfrac12+\sqrt3,\,-3\right)\approx(2.23,-3),\qquad\left(\tfrac12-\sqrt3,\,-3\right)\approx(-1.23,-3)

Dashed lines: the vertical tangents at the two vertices.

\left(\tfrac12+\sqrt3,\,-3\right),\ \left(\tfrac12-\sqrt3,\,-3\right)

(ii) \dfrac{(y-1)^{2}}{2^{2}}-\dfrac{(x+1)^{2}}{19}=1

Differentiate with respect to x.
\dfrac{2(y-1)}{4}\dfrac{dy}{dx}-\dfrac{2(x+1)}{19}=0\;\Rightarrow\;\dfrac{dy}{dx}=\dfrac{4(x+1)}{19(y-1)}
Horizontal tangent: numerator zero, x=-1. Put this in the equation.
\dfrac{(y-1)^{2}}{4}=1\;\Rightarrow\;y-1=\pm2\;\Rightarrow\;y=3\ \text{ or }\ y=-1
Vertical tangent: denominator zero, y=1. Put this in the equation.
-\dfrac{(x+1)^{2}}{19}=1\;\Rightarrow\;(x+1)^{2}=-19
No real x, so there is no vertical tangent. The horizontal tangents y=3 and y=-1 touch the curve at the vertices.
(-1,3),\qquad(-1,-1)

Dashed lines: the horizontal tangents y=3 and y=-1 at the vertices.

(-1,3),\ (-1,-1)

(iii) 5x^{2}-5y^{2}+25x-5y+20=0

Differentiate each term with respect to x.
10x-10y\dfrac{dy}{dx}+25-5\dfrac{dy}{dx}=0\;\Rightarrow\;\dfrac{dy}{dx}=\dfrac{10x+25}{10y+5}=\dfrac{2x+5}{2y+1}
Horizontal tangent: 2x+5=0, so x=-\frac52. Put this in the equation.
5\cdot\dfrac{25}{4}-5y^{2}-\dfrac{125}{2}-5y+20=0\;\Rightarrow\;-5y^{2}-5y-\dfrac{45}{4}=0\;\Rightarrow\;y^{2}+y+\dfrac94=0
Its discriminant is 1-4\left(\frac94\right)=-8<0, so there is no real point and no horizontal tangent.
Vertical tangent: 2y+1=0, so y=-\frac12. Put this in the equation and multiply by \frac45.
5x^{2}-\dfrac54+25x+\dfrac52+20=0\;\Rightarrow\;5x^{2}+25x+\dfrac{85}{4}=0\;\Rightarrow\;4x^{2}+20x+17=0
Solve with the quadratic formula.
x=\dfrac{-20\pm\sqrt{400-272}}{8}=\dfrac{-20\pm8\sqrt2}{8}=-\dfrac52\pm\sqrt2
The vertical tangents touch the curve at the vertices.
\left(-\tfrac52+\sqrt2,\,-\tfrac12\right)\approx(-1.09,-0.5),\qquad\left(-\tfrac52-\sqrt2,\,-\tfrac12\right)\approx(-3.91,-0.5)

Vertices V_{1,2}=\left(-\frac52\mp\sqrt2,-\frac12\right) with their vertical tangents (dashed).

\left(-\tfrac52+\sqrt2,\,-\tfrac12\right),\ \left(-\tfrac52-\sqrt2,\,-\tfrac12\right)

(iv) -x^{2}+y^{2}-10x-4y-28=0

Differentiate each term with respect to x.
-2x+2y\dfrac{dy}{dx}-10-4\dfrac{dy}{dx}=0\;\Rightarrow\;\dfrac{dy}{dx}=\dfrac{2x+10}{2y-4}=\dfrac{x+5}{y-2}
Horizontal tangent: x+5=0, so x=-5. Put this in the equation.
-25+y^{2}+50-4y-28=0\;\Rightarrow\;y^{2}-4y-3=0\;\Rightarrow\;y=\dfrac{4\pm\sqrt{16+12}}{2}=2\pm\sqrt7
Vertical tangent: y-2=0, so y=2. Put this in the equation.
-x^{2}+4-10x-8-28=0\;\Rightarrow\;x^{2}+10x+32=0
Its discriminant is 100-128<0, so there is no vertical tangent. The horizontal tangents y=2\pm\sqrt7 touch the curve at the vertices.
\left(-5,\,2+\sqrt7\right)\approx(-5,4.65),\qquad\left(-5,\,2-\sqrt7\right)\approx(-5,-0.65)

Dashed lines: the horizontal tangents y=2\pm\sqrt7 at the vertices.

\left(-5,\,2+\sqrt7\right),\ \left(-5,\,2-\sqrt7\right)

Exercise 7.9p. 223

Tangent and normal to a hyperbola (Sections 7.13 and 7.14). The methods are the same as for the ellipse: substitute the line and set the discriminant to zero for tangency, or differentiate implicitly for the tangent at a given point. For \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1 the tangent of slope m is y=mx\pm\sqrt{a^{2}m^{2}-b^{2}} and the tangent at (x_{1},y_{1}) is \frac{x_{1}x}{a^{2}}-\frac{y_{1}y}{b^{2}}=1.

Find the length of chord intercepted by the line 2x+y-1=0 and the hyperbola 2x^{2}-3y^{2}+7x-4y+13=0.

Solution

Make y the subject of the line.
2x+y-1=0\;\Rightarrow\;y=1-2x
Substitute in the hyperbola; the end points of the chord satisfy both equations.
2x^{2}-3(1-2x)^{2}+7x-4(1-2x)+13=0
Expand and collect like terms, then multiply by -1.
2x^{2}-3+12x-12x^{2}+7x-4+8x+13=0\;\Rightarrow\;10x^{2}-27x-6=0
The roots x_{1},x_{2} are the abscissae of the end points A and B. Solve with the quadratic formula.
x=\dfrac{27\pm\sqrt{729+240}}{20}=\dfrac{27\pm\sqrt{969}}{20}
So x_{1}-x_{2}=\dfrac{2\sqrt{969}}{20}=\dfrac{\sqrt{969}}{10}. On the line y=1-2x, so y_{1}-y_{2}=-2(x_{1}-x_{2}). Use the distance formula.
|AB|=\sqrt{(x_{1}-x_{2})^{2}+(y_{1}-y_{2})^{2}}=\sqrt{1+4}\;|x_{1}-x_{2}|=\sqrt5\cdot\dfrac{\sqrt{969}}{10}
Simplify.
|AB|=\dfrac{\sqrt{4845}}{10}\approx6.96\ \text{units}
For reference, the end points are:
\begin{aligned}A\left(\dfrac{27+\sqrt{969}}{20},\,\dfrac{-17-\sqrt{969}}{10}\right)&\approx(2.91,-4.81)\\[4pt] B\left(\dfrac{27-\sqrt{969}}{20},\,\dfrac{-17+\sqrt{969}}{10}\right)&\approx(-0.21,1.41)\end{aligned}

The line meets the lower branch at A and the upper branch at B; |AB|\approx6.96.

|AB|=\dfrac{\sqrt{4845}}{10}\approx6.96

Find the value of m so that the line 3x+4y+m=0 is tangent to the hyperbola x^{2}-y^{2}-7x-2y+13=0.

Solution

Make y the subject of the line.
y=-\dfrac{3x+m}{4}
Substitute in the hyperbola and multiply by 16. Note y^{2}=\frac{(3x+m)^{2}}{16} and -2y=\frac{3x+m}{2}.
16x^{2}-(3x+m)^{2}-112x+8(3x+m)+208=0
Expand and collect the powers of x.
16x^{2}-9x^{2}-6mx-m^{2}-112x+24x+8m+208=0\;\Rightarrow\;7x^{2}-(6m+88)x+(208+8m-m^{2})=0
The line is a tangent when this quadratic has equal roots, that is when its discriminant is zero.
(6m+88)^{2}-4(7)(208+8m-m^{2})=0
Expand and simplify; every coefficient is a multiple of 64.
36m^{2}+1056m+7744-5824-224m+28m^{2}=0\;\Rightarrow\;64m^{2}+832m+1920=0\;\Rightarrow\;m^{2}+13m+30=0
Factorise.
(m+3)(m+10)=0\;\Rightarrow\;m=-3\ \text{ or }\ m=-10
Check the points of contact. For m=-3 the quadratic is 7x^{2}-70x+175=7(x-5)^{2}, a double root x=5, point (5,-3). For m=-10 it is 7x^{2}-28x+28=7(x-2)^{2}, point (2,1). Both points satisfy the equation of the hyperbola.

In standard form \dfrac{(y+1)^{2}}{7/4}-\dfrac{(x-\frac72)^{2}}{7/4}=1. The two parallel lines touch different branches.

m=-3\quad\text{or}\quad m=-10

Find the equation of tangent to the hyperbola x^{2}-2y^{2}+4x-6y+11=0 which is parallel to the line 4x-8y+7=0.

Solution

The line 4x-8y+7=0 has slope -\frac{4}{-8}=\frac12. A parallel tangent has the same slope.
y=\dfrac12x+c
Substitute in the hyperbola.
x^{2}-2\left(\dfrac x2+c\right)^{2}+4x-6\left(\dfrac x2+c\right)+11=0
Expand and collect, then multiply by 2.
\dfrac{x^{2}}{2}+(1-2c)x+(11-6c-2c^{2})=0\;\Rightarrow\;x^{2}+2(1-2c)x+2(11-6c-2c^{2})=0
Tangency: the discriminant is zero. Divide it by 4.
4(1-2c)^{2}-8(11-6c-2c^{2})=0\;\Rightarrow\;(1-2c)^{2}-2(11-6c-2c^{2})=0
Simplify.
1-4c+4c^{2}-22+12c+4c^{2}=0\;\Rightarrow\;8c^{2}+8c-21=0
Solve with the quadratic formula; \sqrt{736}=\sqrt{16\times46}=4\sqrt{46}.
c=\dfrac{-8\pm\sqrt{64+672}}{16}=\dfrac{-8\pm4\sqrt{46}}{16}=\dfrac{-2\pm\sqrt{46}}{4}
Put c in y=\frac x2+c and multiply by 4.
4y=2x-2\pm\sqrt{46}\;\Rightarrow\;2x-4y-2\pm\sqrt{46}=0
Points of contact (double root x=-(1-2c)=2c-1, then y=\frac x2+c).
\begin{aligned}\left(-2+\tfrac{\sqrt{46}}{2},\,-\tfrac32+\tfrac{\sqrt{46}}{2}\right)&\approx(1.39,1.89)\\ \left(-2-\tfrac{\sqrt{46}}{2},\,-\tfrac32-\tfrac{\sqrt{46}}{2}\right)&\approx(-5.39,-4.89)\end{aligned}

Both tangents are parallel to the dotted line 4x-8y+7=0.

2x-4y-2+\sqrt{46}=0\quad\text{and}\quad2x-4y-2-\sqrt{46}=0

Find the equations of tangent to the hyperbola 2x^{2}-4y^{2}+6x-8y-7=0 which is perpendicular to the line x+2y+5=0. Also find the point of tangency.

Solution

The line x+2y+5=0 has slope -\frac12. A perpendicular line has slope m with m\left(-\frac12\right)=-1.
m=2\;\Rightarrow\;y=2x+c
Substitute in the hyperbola.
2x^{2}-4(2x+c)^{2}+6x-8(2x+c)-7=0
Expand and collect, then multiply by -1.
2x^{2}-16x^{2}-16cx-4c^{2}+6x-16x-8c-7=0\;\Rightarrow\;14x^{2}+(16c+10)x+(4c^{2}+8c+7)=0
Tangency: the discriminant is zero.
(16c+10)^{2}-4(14)(4c^{2}+8c+7)=0
Expand, simplify and divide by 4.
256c^{2}+320c+100-224c^{2}-448c-392=0\;\Rightarrow\;32c^{2}-128c-292=0\;\Rightarrow\;8c^{2}-32c-73=0
Solve; \sqrt{3360}=\sqrt{16\times210}=4\sqrt{210}.
c=\dfrac{32\pm\sqrt{1024+2336}}{16}=\dfrac{32\pm4\sqrt{210}}{16}=2\pm\dfrac{\sqrt{210}}{4}
Put c in y=2x+c and multiply by 4.
4y=8x+8\pm\sqrt{210}\;\Rightarrow\;8x-4y+8\pm\sqrt{210}=0
Point of tangency. At a tangent the quadratic has the double root x=-\dfrac{16c+10}{2(14)}=-\dfrac{8c+5}{14}; then y=2x+c.
8c+5=21\pm2\sqrt{210}\;\Rightarrow\;x=-\dfrac32\mp\dfrac{\sqrt{210}}{7},\qquad y=-1\mp\dfrac{\sqrt{210}}{28}
So the tangents and their points of contact are:
\begin{aligned}8x-4y+8+\sqrt{210}=0&:\ \left(-\tfrac32-\tfrac{\sqrt{210}}{7},\,-1-\tfrac{\sqrt{210}}{28}\right)\approx(-3.57,-1.52)\\8x-4y+8-\sqrt{210}=0&:\ \left(-\tfrac32+\tfrac{\sqrt{210}}{7},\,-1+\tfrac{\sqrt{210}}{28}\right)\approx(0.57,-0.48)\end{aligned}

The tangents of slope 2 touch the hyperbola at about (-3.57,-1.52) and (0.57,-0.48); both are perpendicular to the dotted line.

8x-4y+8\pm\sqrt{210}=0,\quad\text{points }\left(-\tfrac32\mp\tfrac{\sqrt{210}}{7},\,-1\mp\tfrac{\sqrt{210}}{28}\right)

Find the equations of the tangent and normal to the hyperbola 6x^{2}-6y^{2}-14x+21y-5=0 at the point (2,3).

Solution

Check that (2,3) is on the hyperbola.
6(4)-6(9)-14(2)+21(3)-5=24-54-28+63-5=0
Differentiate implicitly with respect to x (the method of Example 31).
12x-12y\dfrac{dy}{dx}-14+21\dfrac{dy}{dx}=0\;\Rightarrow\;\dfrac{dy}{dx}=\dfrac{12x-14}{12y-21}
Slope of the tangent at (2,3).
m=\dfrac{24-14}{36-21}=\dfrac{10}{15}=\dfrac23
Tangent by the point-slope form.
y-3=\dfrac23(x-2)\;\Rightarrow\;3y-9=2x-4\;\Rightarrow\;2x-3y+5=0
The normal is perpendicular to the tangent, so its slope is -\frac32.
y-3=-\dfrac32(x-2)\;\Rightarrow\;2y-6=-3x+6\;\Rightarrow\;3x+2y-12=0

In standard form \left(y-\frac74\right)^{2}-\left(x-\frac76\right)^{2}=\frac{125}{144}; (2,3) is on the upper branch.

\text{Tangent: }2x-3y+5=0,\qquad\text{Normal: }3x+2y-12=0

Find the equations of the tangents and normal to the hyperbola \dfrac{\left(y-\frac74\right)^{2}}{\frac{25}{48}}-\dfrac{\left(x-\frac76\right)^{2}}{\frac{25}{72}}=1 at the point whose ordinate is 3 and abscissa is an integer.

Solution

Put the ordinate y=3 in the equation. Here 3-\frac74=\frac54 and \dfrac{25/16}{25/48}=3.
3-\dfrac{\left(x-\frac76\right)^{2}}{25/72}=1\;\Rightarrow\;\left(x-\dfrac76\right)^{2}=\dfrac{2\times25}{72}=\dfrac{25}{36}
Take square roots.
x-\dfrac76=\pm\dfrac56\;\Rightarrow\;x=2\ \text{ or }\ x=\dfrac13
The abscissa must be an integer, so the point is (2,3).
Differentiate the equation implicitly with respect to x.
\dfrac{2\left(y-\frac74\right)}{25/48}\dfrac{dy}{dx}-\dfrac{2\left(x-\frac76\right)}{25/72}=0\;\Rightarrow\;\dfrac{dy}{dx}=\dfrac{72\left(x-\frac76\right)}{48\left(y-\frac74\right)}=\dfrac{3\left(x-\frac76\right)}{2\left(y-\frac74\right)}
Slope of the tangent at (2,3).
m=\dfrac{3\left(\frac56\right)}{2\left(\frac54\right)}=\dfrac{5/2}{5/2}=1
Tangent.
y-3=1(x-2)\;\Rightarrow\;x-y+1=0
Normal (slope -1).
y-3=-1(x-2)\;\Rightarrow\;x+y-5=0

Two points of the curve have ordinate 3: (2,3) and \left(\frac13,3\right) (open dot). Only (2,3) has an integer abscissa.

\text{Tangent: }x-y+1=0,\qquad\text{Normal: }x+y-5=0

Prove that the tangent at any point on the hyperbola makes equal angles with the lines joining the point with the foci of the hyperbola.

Solution

Idea. This is the reflecting property of the hyperbola: the tangent at P bisects the angle between the focal lines PF_{1} and PF_{2}. As for the ellipse (Exercise 7.7, Q.7) we compare the angles with \tan\theta=\dfrac{m_{2}-m_{1}}{1+m_{1}m_{2}}, the angle turned from the line of slope m_{1} to the line of slope m_{2}.

Take the hyperbola, its foci and any point P(x_{1},y_{1}) on it with y_{1}\neq0.
\dfrac{x^{2}}{a^{2}}-\dfrac{y^{2}}{b^{2}}=1,\qquad F_{1}(-c,0),\ F_{2}(c,0),\qquad c^{2}=a^{2}+b^{2}
P is on the curve; multiply by a^{2}b^{2}.
b^{2}x_{1}^{2}-a^{2}y_{1}^{2}=a^{2}b^{2}\qquad(1)
Slope of the tangent at P (differentiate b^{2}x^{2}-a^{2}y^{2}=a^{2}b^{2} implicitly).
2b^{2}x-2a^{2}y\dfrac{dy}{dx}=0\;\Rightarrow\;m=\dfrac{b^{2}x_{1}}{a^{2}y_{1}}
Slopes of PF_{1} and PF_{2}.
m_{1}=\dfrac{y_{1}}{x_{1}+c},\qquad m_{2}=\dfrac{y_{1}}{x_{1}-c}
Angle \alpha from PF_{1} to the tangent. Simplify the numerator of \tan\alpha=\dfrac{m-m_{1}}{1+mm_{1}} with (1).
m-m_{1}=\dfrac{b^{2}x_{1}(x_{1}+c)-a^{2}y_{1}^{2}}{a^{2}y_{1}(x_{1}+c)}=\dfrac{a^{2}b^{2}+b^{2}cx_{1}}{a^{2}y_{1}(x_{1}+c)}=\dfrac{b^{2}(a^{2}+cx_{1})}{a^{2}y_{1}(x_{1}+c)}
Simplify the denominator with a^{2}+b^{2}=c^{2}.
1+mm_{1}=\dfrac{a^{2}(x_{1}+c)+b^{2}x_{1}}{a^{2}(x_{1}+c)}=\dfrac{c^{2}x_{1}+a^{2}c}{a^{2}(x_{1}+c)}=\dfrac{c(a^{2}+cx_{1})}{a^{2}(x_{1}+c)}
Divide; the common factors cancel.
\tan\alpha=\dfrac{b^{2}}{cy_{1}}
Angle \beta from the tangent to PF_{2}, \tan\beta=\dfrac{m_{2}-m}{1+mm_{2}}. Work in the same way.
\begin{aligned}m_{2}-m&=\dfrac{a^{2}y_{1}^{2}-b^{2}x_{1}(x_{1}-c)}{a^{2}y_{1}(x_{1}-c)}=\dfrac{b^{2}(cx_{1}-a^{2})}{a^{2}y_{1}(x_{1}-c)}\\1+mm_{2}&=\dfrac{a^{2}(x_{1}-c)+b^{2}x_{1}}{a^{2}(x_{1}-c)}=\dfrac{c(cx_{1}-a^{2})}{a^{2}(x_{1}-c)}\end{aligned}
Divide (cx_{1}-a^{2}\neq0 because |x_{1}|\ge a and c>a).
\tan\beta=\dfrac{b^{2}}{cy_{1}}
So \tan\alpha=\tan\beta and therefore \alpha=\beta.
\alpha=\beta

The tangent at P makes equal angles with PF_{1} and PF_{2}, so it bisects \angle F_{1}PF_{2}. A ray of light aimed at one focus is reflected by a hyperbolic mirror towards the other focus. Hence proved.

Example with a=3, b=4, c=5 and P(3.75,\,3): the tangent lies between PF_{1} and PF_{2} and makes equal angles (about 47^\circ, since \tan\alpha=\frac{b^{2}}{cy_{1}}=\frac{16}{15}) with them.

\tan\alpha=\tan\beta=\dfrac{b^{2}}{cy_{1}}\;\Rightarrow\;\alpha=\beta

The line 12x-5\sqrt{11}-50=0 is tangent to the hyperbola \dfrac{x^{2}}{25}-\dfrac{y^{2}}{4}=1. Find the point where the tangent line intersects the focal axis of hyperbola. Also find the ratio in which this point of intersection divides the distance between the foci.

Solution

Find where the line 12x-5\sqrt{11}\,y-50=0 touches the curve. The tangent at (x_{1},y_{1}) is \frac{x_{1}x}{25}-\frac{y_{1}y}{4}=1; divide the line by 50 and compare coefficients.
\dfrac{12x}{50}-\dfrac{\sqrt{11}\,y}{10}=1\;\Rightarrow\;\dfrac{x_{1}}{25}=\dfrac{12}{50},\ \dfrac{y_{1}}{4}=\dfrac{\sqrt{11}}{10}\;\Rightarrow\;x_{1}=6,\ y_{1}=\dfrac{2\sqrt{11}}{5}
Check that P\left(6,\frac{2\sqrt{11}}{5}\right) is on the hyperbola.
\dfrac{36}{25}-\dfrac{44/25}{4}=\dfrac{36}{25}-\dfrac{11}{25}=1
The focal axis of \frac{x^{2}}{25}-\frac{y^{2}}{4}=1 is the x-axis, y=0. Put y=0 in the tangent.
12x-50=0\;\Rightarrow\;x=\dfrac{25}{6}\;\Rightarrow\;T\left(\dfrac{25}{6},\,0\right)
Foci: a^{2}=25, b^{2}=4, so c^{2}=a^{2}+b^{2}=29.
F_{1}(-\sqrt{29},0),\qquad F_{2}(\sqrt{29},0),\qquad\sqrt{29}\approx5.385
T lies between the foci because \frac{25}{6}\approx4.17<\sqrt{29}. Measure the two parts along the axis.
|F_{1}T|=\dfrac{25}{6}+\sqrt{29}=\dfrac{25+6\sqrt{29}}{6},\qquad|TF_{2}|=\sqrt{29}-\dfrac{25}{6}=\dfrac{6\sqrt{29}-25}{6}
So T divides F_{1}F_{2} internally in the ratio:
|F_{1}T|:|TF_{2}|=(6\sqrt{29}+25):(6\sqrt{29}-25)\approx9.552:1.219\approx7.84:1
Check with Q.7: the tangent bisects \angle F_{1}PF_{2}, so by the angle bisector theorem T divides F_{1}F_{2} in the ratio |PF_{1}|:|PF_{2}|. By the distance formula, |PF_{1}|^{2}=(6+\sqrt{29})^{2}+\frac{44}{25}=\left(\frac{6\sqrt{29}}{5}+5\right)^{2} and |PF_{2}|^{2}=(6-\sqrt{29})^{2}+\frac{44}{25}=\left(\frac{6\sqrt{29}}{5}-5\right)^{2}.
|PF_{1}|:|PF_{2}|=\left(\dfrac{6\sqrt{29}+25}{5}\right):\left(\dfrac{6\sqrt{29}-25}{5}\right)=(6\sqrt{29}+25):(6\sqrt{29}-25)

The tangent at P\left(6,\frac{2\sqrt{11}}{5}\right) meets the focal axis at T\left(\frac{25}{6},0\right), between the foci and much closer to F_{2}.

T\left(\tfrac{25}{6},0\right),\qquad F_{1}T:TF_{2}=(6\sqrt{29}+25):(6\sqrt{29}-25)\approx7.84:1

Note. Misprint: the line has lost its y. The tangent is 12x-5\sqrt{11}\,y-50=0; the printed 12x-5\sqrt{11}-50=0 is only a vertical line and does not touch the hyperbola.

Exercise 7.10p. 227–228

Applications of conic sections (Sections 7.15 and 7.16). Choose axes that make the equation simple (centre or vertex at the origin), turn every measurement into the coordinates of a point, find the equation, then answer the question in words with units. Remember the reflecting properties: a parabola sends parallel rays to its focus, an ellipse sends rays from one focus to the other.

A cell phone company has installed three signal towers A, B and C at different locations to provide service to their customers. Tower A is located at the position (-2,10) and covers an area up to 8\,\mathrm{km}, tower B is located at (-3,5) and covers an area up to 6\,\mathrm{km} and tower C is located at (10,-1) and covers an area up to 5\,\mathrm{km}. A man is standing at a position of (12,3), which tower will provide service to the man?

Solution

Each tower covers a circular region: the tower is the centre and its range is the radius (take 1 unit =1\,\mathrm{km}). The man M(12,3) gets service from a tower when his distance from it is not more than its radius.
Distance from tower A(-2,10), radius 8.
|MA|=\sqrt{(12+2)^{2}+(3-10)^{2}}=\sqrt{196+49}=\sqrt{245}\approx15.65\,\mathrm{km}>8\,\mathrm{km}
Distance from tower B(-3,5), radius 6.
|MB|=\sqrt{(12+3)^{2}+(3-5)^{2}}=\sqrt{225+4}=\sqrt{229}\approx15.13\,\mathrm{km}>6\,\mathrm{km}
Distance from tower C(10,-1), radius 5.
|MC|=\sqrt{(12-10)^{2}+(3+1)^{2}}=\sqrt{4+16}=\sqrt{20}\approx4.47\,\mathrm{km}<5\,\mathrm{km}
The same test with the equation of circle C, (x-10)^{2}+(y+1)^{2}=25: put M(12,3) in the left side minus 25. A negative value means the point is inside the circle.
(12-10)^{2}+(3+1)^{2}-25=4+16-25=-5<0
M is outside the circles of A and B but inside the circle of C. So only tower C provides service to the man.

Coverage areas of the three towers. The man M lies only inside circle C.

\begin{aligned}&\text{Tower C}\\ &(|MC|=2\sqrt5\approx4.47\,\mathrm{km}<5\,\mathrm{km})\end{aligned}

The tyre of a car is of radius 15 inches and it revolves 1000 times in a minute. What is the speed of the car in \frac{\mathrm{km}}{\mathrm{h}}?

Solution

The tyre rolls without slipping, so in one revolution the car moves forward one circumference of the tyre (a circle of radius r=15 inches).
2\pi r=2\pi(15)=30\pi\ \text{inches}
Distance moved in one minute (1000 revolutions).
1000\times30\pi=30000\pi\ \text{inches}
Convert to kilometres with 1\ \text{inch}=2.54\,\mathrm{cm} and 1\,\mathrm{km}=100000\,\mathrm{cm}.
30000\pi\times2.54\,\mathrm{cm}=76200\pi\,\mathrm{cm}=\dfrac{76200\pi}{100000}\,\mathrm{km}=0.762\pi\,\mathrm{km}
This is the distance per minute. There are 60 minutes in an hour.
v=0.762\pi\times60=45.72\pi\approx143.6\,\mathrm{km\,h^{-1}}

v=45.72\pi\approx143.6\,\mathrm{km\,h^{-1}}

A batsman hits the ball. The ball attains the maximum height of 25\,\mathrm{m} and drops on the ground at a distance of 80\,\mathrm{m} from the batsman. Assuming the origin at the position of batsman write the equation of path of the ball. Is it possible for a man of height 1.6\,\mathrm{m} to catch the ball, standing 70\,\mathrm{m} away from the batsman?

Solution

The path of the ball is a parabola opening downwards. Take the origin at the batsman and the x-axis along the ground. The ball lands at (80,0), and by symmetry its highest point (the vertex) is halfway, at (40,25).
A parabola opening downwards with vertex (h,k) has the equation below.
(x-h)^{2}=-4a(y-k)\;\Rightarrow\;(x-40)^{2}=-4a(y-25)
The ball starts at O(0,0), which lies on the path.
(0-40)^{2}=-4a(0-25)\;\Rightarrow\;1600=100a\;\Rightarrow\;a=16
Equation of the path (it also passes through (80,0)).
(x-40)^{2}=-64(y-25)\quad\text{or}\quad y=\dfrac{x(80-x)}{64}
Height of the ball at 70\,\mathrm{m} from the batsman: put x=70.
y=\dfrac{70(80-70)}{64}=\dfrac{700}{64}\approx10.94\,\mathrm{m}
At that position the ball is about 10.94\,\mathrm{m} above the ground, far above the reach of a man 1.6\,\mathrm{m} tall. So it is not possible for him to catch the ball there.

Equal scales on both axes. At x=70 the ball is still about 10.94\,\mathrm{m} high, while the man reaches only 1.6\,\mathrm{m}.

(x-40)^{2}=-64(y-25);\ \ y(70)\approx10.94\,\mathrm{m}>1.6\,\mathrm{m},\ \text{so he cannot catch it}

Playing team coaches use parabolic shaped antennas to listen the conversation between the players in the ground. If the antenna has the cross section of 18 inches and depth of 4 inches then where the microphone should be placed to hear the conversation?

Solution

Sound arriving parallel to the axis of a parabolic dish is reflected to the focus, so the microphone must be placed at the focus. Place the vertex of the cross-section at the origin with the axis along the y-axis (dish opening upwards).
x^{2}=4ay,\qquad\text{focus }F(0,a)
The opening is 18 inches wide at a depth of 4 inches, so the rim point (9,4) is on the parabola (half of 18 is 9).
9^{2}=4a(4)\;\Rightarrow\;81=16a\;\Rightarrow\;a=\dfrac{81}{16}=5.0625
So the focus is on the axis, \frac{81}{16} inches from the vertex.
F\left(0,\dfrac{81}{16}\right)\approx(0,\,5.06)
The microphone should be placed on the axis of the antenna about 5.06 inches from its vertex (centre of the dish). The dish is only 4 inches deep, so this point is about 1.06 inches beyond the plane of the rim.

Cross-section of the antenna. The microphone goes at the focus F, 5.06 inches from the vertex.

\text{At the focus: on the axis, } a=\dfrac{81}{16}\approx5.06\ \text{inches from the vertex}

Man A is standing at one focus of the whispering gallery which is 16 feet from the nearest wall. The man B is standing at the other focus 100\,\mathrm{ft} away. What is the length of the whispering gallery? How high is the elliptical ceiling at the centre?

Solution

A whispering gallery has an elliptical ceiling: a whisper from one focus reflects to the other focus. Place the centre of the ellipse at the origin with the foci on the x-axis.
\dfrac{x^{2}}{a^{2}}+\dfrac{y^{2}}{b^{2}}=1,\qquad c^{2}=a^{2}-b^{2}
The two men are at the foci, 100\,\mathrm{ft} apart.
2c=100\;\Rightarrow\;c=50
The nearest wall is at the end of the major axis on A's side, so the distance from a focus to it is a-c.
a-c=16\;\Rightarrow\;a=66
The length of the gallery is the major axis.
2a=132\,\mathrm{ft}
The height of the ceiling at the centre is the semi-minor axis b.
b=\sqrt{a^{2}-c^{2}}=\sqrt{4356-2500}=\sqrt{1856}=8\sqrt{29}\approx43.1\,\mathrm{ft}

The gallery is 132\,\mathrm{ft} long; each focus is 16\,\mathrm{ft} from the nearer end wall; the ceiling is about 43.1\,\mathrm{ft} high at the centre.

\text{Length}=132\,\mathrm{ft},\qquad\text{height at centre}=8\sqrt{29}\approx43.1\,\mathrm{ft}

The orbit of a planet is in elliptical shape with Sun setting at one of its two foci. The eccentricity of the ellipse is 0.085 and the minimum distance of the planet from the Sun is 105 million miles. What is the maximum distance of the planet from Sun?

Solution

Take the orbit \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1 with the Sun at the focus F(c,0), where c=ae (as in Example 35).
The planet is nearest to the Sun at the vertex (a,0) on the Sun's side and farthest at the other vertex (-a,0).
r_{\min}=a-c=a(1-e),\qquad r_{\max}=a+c=a(1+e)
Use the minimum distance to find a (distances in million miles).
a(1-0.085)=105\;\Rightarrow\;a=\dfrac{105}{0.915}\approx114.75
Maximum distance.
r_{\max}=a(1+e)=114.75\times1.085\approx124.5
Shortcut: dividing the two formulas gives \dfrac{r_{\max}}{r_{\min}}=\dfrac{1+e}{1-e}.
r_{\max}=105\times\dfrac{1.085}{0.915}\approx124.5\ \text{million miles}

With e=0.085 the orbit is almost a circle; the Sun is only c=ae\approx9.75 million miles from the centre.

r_{\max}=105\times\dfrac{1.085}{0.915}\approx124.5\ \text{million miles}

The control towers are located at points Q(-500,0) and R(500,0) on a straight shore where the x-axis runs through. At same moment both towers send a radio signal to a ship out at sea, each travelling at 300\,\mathrm{m}/\mu\mathrm{s}. The ship received the signal from Q, 3\,\mu\mathrm{s} earlier the message from R. Find the equation of hyperbola containing the possible location of the ship.

Solution

Both signals travel at 300\,\mathrm{m} per microsecond. The signal from R arrives 3\,\mu\mathrm{s} later, so the ship S is farther from R than from Q by:
|SR|-|SQ|=300\times3=900\,\mathrm{m}
The difference of the distances from the two fixed points Q and R is constant, so S lies on a hyperbola with foci Q and R, and this difference is 2a.
2a=900\;\Rightarrow\;a=450
The foci (\pm500,0) give the centre (0,0), c=500 and a horizontal focal axis.
\dfrac{x^{2}}{a^{2}}-\dfrac{y^{2}}{b^{2}}=1
Find b^{2}.
b^{2}=c^{2}-a^{2}=500^{2}-450^{2}=250000-202500=47500
Equation of the hyperbola (units in metres).
\dfrac{x^{2}}{202500}-\dfrac{y^{2}}{47500}=1
The ship is closer to Q, so it lies on the branch nearer Q, the left branch (x\le-450).

\frac{x^{2}}{202500}-\frac{y^{2}}{47500}=1. Points on the left branch are 900\,\mathrm{m} closer to Q than to R.

\dfrac{x^{2}}{202500}-\dfrac{y^{2}}{47500}=1\quad(\text{branch nearer }Q,\ x\le-450)

An architect's design for a building includes some large pillars with cross sections in the shape of hyperbolas. The curve can be modelled by the equation \dfrac{x^{2}}{16}-\dfrac{y^{2}}{9}=1 where the units are in meters. If the pillars are 4.2\,\mathrm{m} tall. Find the width of the top of each pillar. Also find the width at the middle of the pillar.

Solution

The two branches of the hyperbola form the curved sides of the pillar. The centre of the hyperbola is at the middle of the pillar, so a pillar 4.2\,\mathrm{m} tall runs from y=-2.1 to y=2.1.
At height y the width is the distance between the branches, 2x. Solve the equation for x^{2}.
x^{2}=16\left(1+\dfrac{y^{2}}{9}\right)
Top of the pillar: y=2.1.
x^{2}=16\left(1+\dfrac{4.41}{9}\right)=16(1.49)=23.84\;\Rightarrow\;x\approx4.883
Width of the top.
2x\approx2(4.883)\approx9.77\,\mathrm{m}
Middle of the pillar: y=0, where the pillar is narrowest.
x^{2}=16\;\Rightarrow\;x=4\;\Rightarrow\;2x=8\,\mathrm{m}

Cross-section of one pillar from y=-2.1 to y=2.1: widest at the top and bottom, narrowest (8\,\mathrm{m}) in the middle.

\text{Top width}=8\sqrt{1.49}\approx9.77\,\mathrm{m},\qquad\text{middle width}=8\,\mathrm{m}

Review exercisep. 228–229

Question 1 revises the definitions and standard results of the whole unit (circle, parabola, ellipse and hyperbola). Questions 2 to 5 combine them in problems: tangent to a circle, a parabolic arch, focal distances of an ellipse and the reflecting property of a hyperbola.

1 (i)

The eccentricity is the ratio of distance of a point on the conic section from:

AFocus to directrixAnswer
BDirectrix to focus
CVertex to directrix
DDirectrix to vertex

Why

By the focus-directrix definition, e=\dfrac{|PF|}{|PM|}: the distance of the point P from the focus divided by its distance from the directrix. So the ratio is focus to directrix. Option (b) is the reciprocal of e.

1 (ii)

Eccentricity of circle is:

Ae>1
Be<1
Ce=1
De=0Answer

Why

For an ellipse e=\frac ca, where c is the distance from the centre to a focus. A circle is an ellipse whose two foci coincide at the centre, so c=0 and e=0. (An ellipse has e<1, a parabola e=1 and a hyperbola e>1.)

1 (iii)

The focus of the parabola x^{2}=-16y is:

A(4,0)
B(-4,0)
C(0,4)
D(0,-4)Answer

Why

Compare with x^{2}=-4ay: 4a=16, so a=4. This parabola opens downwards along the y-axis and its focus is (0,-a)=(0,-4). Options (a) and (b) would need y^{2} in the equation.

1 (iv)

Length of the latus rectum of the ellipse \dfrac{x^{2}}{16}+\dfrac{y^{2}}{9}=1 is:

A\frac{32}{9}
B\frac{9}{32}
C\frac{9}{2}Answer
D\frac{8}{9}

Why

Here a^{2}=16 (a=4) and b^{2}=9. The latus rectum has length \dfrac{2b^{2}}{a}=\dfrac{2(9)}{4}=\dfrac92. Option (a) comes from using \frac{2a^{2}}{b}, the formula the wrong way round.

1 (v)

Equation of the directrices of ellipse \dfrac{x^{2}}{16}+\dfrac{y^{2}}{36}=1 is:

Ay=\pm\frac{18}{\sqrt5}Answer
By=\pm\frac{\sqrt5}{18}
Cx=\frac{18}{\sqrt5}
Dx=\frac{\sqrt5}{18}

Why

The larger denominator is under y^{2}, so the major axis is the y-axis with a^{2}=36, b^{2}=16. Then c=\sqrt{36-16}=2\sqrt5 and e=\frac ca=\frac{\sqrt5}{3}. The directrices are horizontal lines y=\pm\frac ae=\pm\frac{6\cdot3}{\sqrt5}=\pm\frac{18}{\sqrt5}.

1 (vi)

Eccentricity of the hyperbola \dfrac{x^{2}}{25}-\dfrac{y^{2}}{81}=1 is:

A\frac{5}{\sqrt{106}}
B\frac{\sqrt{106}}{5}Answer
C\frac{\sqrt{106}}{9}
D\frac{9}{\sqrt{106}}

Why

For a hyperbola c^{2}=a^{2}+b^{2}=25+81=106, so e=\frac ca=\frac{\sqrt{106}}{5}\approx2.06. The eccentricity of a hyperbola is greater than 1, which rules out (a) and (d).

1 (vii)

Equation of conjugate axis of the hyperbola \dfrac{(x-1)^{2}}{4}-\dfrac{(y+3)^{2}}{12}=1 is:

Ax=1Answer
Bx=-1
Cy=3
Dy=-3

Why

The centre is (1,-3) and the positive term contains x, so the transverse (focal) axis is the horizontal line y=-3. The conjugate axis is the perpendicular line through the centre, X=0, that is x=1. Option (d) is the transverse axis.

1 (viii)

Length of the tangent drawn from the point (1,2) to the circle 2x^{2}+2y^{2}+3x+2y-6=0 is:

A11
B\sqrt{11}
C\frac{11}{2}
D\sqrt{\frac{11}{2}}Answer

Why

First make the coefficients of x^{2} and y^{2} equal to 1: x^{2}+y^{2}+\frac32x+y-3=0. The length of the tangent from (x_{1},y_{1}) is \sqrt{x_{1}^{2}+y_{1}^{2}+2gx_{1}+2fy_{1}+c}=\sqrt{1+4+\frac32+2-3}=\sqrt{\frac{11}{2}}. Forgetting to divide by 2 gives the wrong option (b).

1 (ix)

The chord joining the two points (at_{1}^{2},2at_{1}) and (at_{2}^{2},2at_{2}) on the parabola y^{2}=4ax is focal chord if:

At_{1}+t_{2}=1
Bt_{1}+t_{2}=-1
Ct_{1}t_{2}=1
Dt_{1}t_{2}=-1Answer

Why

A focal chord passes through the focus F(a,0), so P, F, Q are collinear. Equal slopes of PF and QF give \dfrac{2t_{1}}{t_{1}^{2}-1}=\dfrac{2t_{2}}{t_{2}^{2}-1}, which simplifies to (t_{2}-t_{1})(t_{1}t_{2}+1)=0. Since t_{1}\neq t_{2}, t_{1}t_{2}=-1 (Exercise 7.5, Q.8).

1 (x)

Exactly one tangent can be drawn to a circle if point lies:

Aoutside the circle
Bon the circleAnswer
Cinside circle
Dcentre of circle

Why

From a point outside the circle two tangents can be drawn, from a point on the circle exactly one (perpendicular to the radius at that point), and from a point inside the circle (including the centre) none.

A point P(a,b) lies outside a circle with centre at origin. From P a tangent is drawn to the circle at point A(2,2\sqrt3). If the area of the triangle POA is 10 sq units then find equation of the circle and the coordinates of the point P.

The book's figure: the tangent AP has length 5.

Solution

The radius is the distance from the centre O(0,0) to the point of contact A.
r=|OA|=\sqrt{2^{2}+(2\sqrt3)^{2}}=\sqrt{4+12}=4
Circle with centre at the origin and radius 4.
x^{2}+y^{2}=16
A tangent is perpendicular to the radius at the point of contact, so \triangle POA has a right angle at A. Its area gives |AP|.
\tfrac12|OA|\,|AP|=10\;\Rightarrow\;\tfrac12(4)|AP|=10\;\Rightarrow\;|AP|=5
By Pythagoras in \triangle POA, |OP|^{2}=|OA|^{2}+|AP|^{2}.
a^{2}+b^{2}=16+25=41\qquad(1)
P lies on the tangent at A(2,2\sqrt3), whose equation is xx_{1}+yy_{1}=r^{2}.
2a+2\sqrt3\,b=16\;\Rightarrow\;a=8-\sqrt3\,b\qquad(2)
Substitute (2) in (1) and simplify.
(8-\sqrt3\,b)^{2}+b^{2}=41\;\Rightarrow\;64-16\sqrt3\,b+4b^{2}=41\;\Rightarrow\;4b^{2}-16\sqrt3\,b+23=0
Solve with the quadratic formula.
b=\dfrac{16\sqrt3\pm\sqrt{768-368}}{8}=\dfrac{16\sqrt3\pm20}{8}=2\sqrt3\pm\dfrac52
Find a from (2) for each value of b.
\begin{aligned}b&=2\sqrt3-\tfrac52\\ \Rightarrow a&=8-6+\tfrac{5\sqrt3}{2}=2+\tfrac{5\sqrt3}{2}\\[6pt] b&=2\sqrt3+\tfrac52\\ \Rightarrow a&=2-\tfrac{5\sqrt3}{2}\end{aligned}
So P can be on either side of A along the tangent.
\begin{aligned}P\left(2+\tfrac{5\sqrt3}{2},\,2\sqrt3-\tfrac52\right)&\approx(6.33,\,0.96)\\ \text{or}\quad P\left(2-\tfrac{5\sqrt3}{2},\,2\sqrt3+\tfrac52\right)&\approx(-2.33,\,5.96)\end{aligned}
Check the first point: area =\frac12|x_{A}y_{P}-y_{A}x_{P}|=\frac12\left|2\left(2\sqrt3-\frac52\right)-2\sqrt3\left(2+\frac{5\sqrt3}{2}\right)\right|=\frac12|-5-15|=10.

P_{1}\approx(6.33,0.96) and P_{2}\approx(-2.33,5.96) are both 5 units from A on the tangent; the shaded triangle OAP_{1} has area 10.

x^{2}+y^{2}=16;\qquad P\left(2+\tfrac{5\sqrt3}{2},\,2\sqrt3-\tfrac52\right)\ \text{or}\ P\left(2-\tfrac{5\sqrt3}{2},\,2\sqrt3+\tfrac52\right)

On the Siri Nagar highway overhead steel bridges are constructed to cross the road. They are supported by the parabolic arc over them. The highway is 300\,\mathrm{ft} wide and walkway is 15\,\mathrm{ft} above the highway. The centre of the parabolic arc is 10\,\mathrm{ft} high from the walkway. Find the equation of the parabolic arc take origin at the mid of the road.

Solution

Take the origin at the middle of the road, the x-axis across the road and the y-axis upwards. The walkway spans the 300\,\mathrm{ft} width at a height of 15\,\mathrm{ft}, so its ends are at (\pm150,15).
The arc stands on the ends of the walkway, and its centre (highest point) is 10\,\mathrm{ft} above the walkway. So the vertex is V(0,25) and the arc passes through (150,15) and (-150,15).
A parabola opening downwards with vertex (0,25) has the form:
(x-0)^{2}=-4a(y-25)
Substitute the end point (150,15).
150^{2}=-4a(15-25)\;\Rightarrow\;22500=40a\;\Rightarrow\;4a=2250
Equation of the parabolic arc.
x^{2}=-2250(y-25)\quad\text{or}\quad y=25-\dfrac{x^{2}}{2250},\qquad-150\le x\le150

The vertical scale is stretched (the real arc is only 10\,\mathrm{ft} high over a 300\,\mathrm{ft} span). The x-axis is the road.

x^{2}=-2250(y-25)

Note. The question does not say where the ends of the arc stand. We take them at the two ends of the walkway, (\pm150,15). If the arc is taken to rise from the edges of the road, (\pm150,0), with the same top (0,25), the same method gives x^{2}=-900(y-25).

If r_{1} and r_{2} are the minimum and maximum distance from a focus of an ellipse then prove that semi-major axis is the arithmetic mean between r_{1} and r_{2} and semi-minor axis is the geometric mean between r_{1} and r_{2}.

Solution

Take the ellipse and one of its foci.
\dfrac{x^{2}}{a^{2}}+\dfrac{y^{2}}{b^{2}}=1\ (a>b),\qquad F(c,0),\quad c^{2}=a^{2}-b^{2},\quad c=ae
Distance from F to any point P(x,y) of the ellipse. From the equation, y^{2}=b^{2}-\frac{b^{2}x^{2}}{a^{2}}.
|PF|^{2}=(x-c)^{2}+y^{2}=x^{2}-2cx+c^{2}+b^{2}-\dfrac{b^{2}x^{2}}{a^{2}}
Use c^{2}+b^{2}=a^{2}, 1-\frac{b^{2}}{a^{2}}=\frac{c^{2}}{a^{2}}=e^{2} and c=ae.
|PF|^{2}=e^{2}x^{2}-2aex+a^{2}=(a-ex)^{2}
Every point of the ellipse has -a\le x\le a and e<1, so a-ex>0 and we can take the square root.
|PF|=a-ex
a-ex is smallest when x=a and largest when x=-a, the two ends of the major axis.
r_{1}=a-ae=a-c,\qquad r_{2}=a+ae=a+c
Arithmetic mean of r_{1} and r_{2}.
\dfrac{r_{1}+r_{2}}{2}=\dfrac{(a-c)+(a+c)}{2}=a
Geometric mean of r_{1} and r_{2}.
\sqrt{r_{1}r_{2}}=\sqrt{(a-c)(a+c)}=\sqrt{a^{2}-c^{2}}=\sqrt{b^{2}}=b

So the semi-major axis a is the arithmetic mean and the semi-minor axis b is the geometric mean of r_{1} and r_{2}. Hence proved.

From the focus F(4,0) the nearest point of the ellipse is (5,0), r_{1}=1, and the farthest is (-5,0), r_{2}=9.

a=\dfrac{r_{1}+r_{2}}{2},\qquad b=\sqrt{r_{1}r_{2}}

A hyperbolic mirror has the property that the light directed at the focus will be reflected to the other focus. If one of the foci has coordinates (24,0) and the top mount point of mirror has coordinates (24,24). Find the vertex of the mirror.

Solution

Take the centre of the hyperbola at the origin and the focal axis along the x-axis, so the foci are (\pm24,0) and c=24.
\dfrac{x^{2}}{a^{2}}-\dfrac{y^{2}}{b^{2}}=1,\qquad a^{2}+b^{2}=c^{2}=576\qquad(1)
The mount point (24,24) is directly above the focus (24,0), so it is an end of the latus rectum. Half the latus rectum is \frac{b^{2}}{a} (put x=c in the equation: y^{2}=b^{2}\cdot\frac{c^{2}-a^{2}}{a^{2}}=\frac{b^{4}}{a^{2}}).
\dfrac{b^{2}}{a}=24\;\Rightarrow\;b^{2}=24a
Substitute in (1).
a^{2}+24a-576=0
Solve with the quadratic formula and keep the positive root (\sqrt{2880}=24\sqrt5).
a=\dfrac{-24+\sqrt{576+2304}}{2}=\dfrac{-24+24\sqrt5}{2}=12(\sqrt5-1)\approx14.83
The mirror is the branch through (24,24), the right branch, so its vertex is (a,0).
V\left(12(\sqrt5-1),\,0\right)\approx(14.83,\,0)
Check: a^{2}\approx220.0, b^{2}=24a\approx356.0, and \frac{24^{2}}{a^{2}}-\frac{24^{2}}{b^{2}}\approx2.618-1.618=1, so (24,24) is on the hyperbola.

The mirror is part of the right branch. The mount point (24,24) is at the end of the latus rectum through F_{2}; the vertex is V\approx(14.83,0).

V\left(12\sqrt5-12,\ 0\right)\approx(14.83,\ 0)

FBISE 2026 questions

Questions from the FBISE HSSC-II 2026 (1st Annual) papers that belong to this unit, answered the way an examiner expects.

2026 Annual I, Set B, MCQ 71 mark

For hyperbola \dfrac{x^2}{16}-\dfrac{y^2}{9}=1, slopes of asymptotes are:

A\pm\dfrac{16}{9}
B\pm\dfrac43
C\pm\dfrac{9}{16}
D\pm\dfrac34Answer

Why

Here a^2=16, b^2=9, so a=4, b=3. The asymptotes of \dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1 are y=\pm\dfrac{b}{a}x=\pm\dfrac34x, so the slopes are \pm\dfrac34.

Option B, \pm\tfrac43, is a/b (upside down); options A and C use a^2 and b^2 instead of a and b.

2026 Annual I, Set B, MCQ 151 mark

The slope of the tangent to the circle x^2+y^2=25 at the point (3,4) is:

A-\dfrac43
B-\dfrac34Answer
C\dfrac34
D\dfrac43

Why

Differentiating x^2+y^2=25 gives \dfrac{dy}{dx}=-\dfrac{x}{y}. At (3,4) the slope is -\dfrac34.

Option D, \tfrac43, is the slope of the radius to (3,4); the tangent is perpendicular to it, so its slope is the negative reciprocal.

2026 Annual I, Set B, MCQ 181 mark

The length of the tangent from point (10,0) to circle x^2+y^2=36 is:

A12
B10
C8Answer
D6

Why

Length of tangent =\sqrt{x_1^2+y_1^2-r^2}=\sqrt{100+0-36}=\sqrt{64}=8. The radius (6), the tangent (8) and the distance to the centre (10) form a 6, 8, 10 right triangle.

2026 Annual I, Set B, Q.2 (iv) OR4 marks

Check whether the line y=2x+1 intersects, touches, or misses the circle x^2+y^2=5.

Solution

The circle x^2+y^2=5 has centre O(0,0) and radius r=\sqrt5\approx2.24.
Write the line as 2x-y+1=0 and find the distance of the centre from it.
d=\dfrac{|2(0)-0+1|}{\sqrt{2^2+(-1)^2}}=\dfrac{1}{\sqrt5}\approx0.45
Compare: d<r, so the line intersects the circle at two points.
\dfrac{1}{\sqrt5}<\sqrt5
Confirm by substitution: put y=2x+1 in the circle.
x^2+(2x+1)^2=5\;\Longrightarrow\;5x^2+4x-4=0,\qquad\Delta=16+80=96>0
Points of intersection.
x=\dfrac{-4\pm\sqrt{96}}{10}=\dfrac{-2\pm2\sqrt6}{5},\qquad y=2x+1=\dfrac{1\pm4\sqrt6}{5}
So the points are approximately (0.58,\,2.16) and (-1.38,\,-1.76).

\text{Intersects at }\left(\tfrac{-2\pm2\sqrt6}{5},\ \tfrac{1\pm4\sqrt6}{5}\right)

2026 Annual I, Set B, Q.2 (vii)4 marks

For the parabola y^2=-12x, find the vertex, focus, directrix, and length of latus rectum.

Solution

Compare with y^2=-4ax, the parabola that opens to the left.
4a=12\;\Longrightarrow\;a=3
Vertex: no shift, so it is the origin.
V(0,0)
Focus: a units to the left of the vertex, on the axis y=0.
F(-a,0)=(-3,0)
Directrix: a units to the right of the vertex, perpendicular to the axis.
x=a\;\Longrightarrow\;x=3\quad(x-3=0)
Length of latus rectum.
|4a|=12
The ends of the latus rectum are (-3,\pm6), since y^2=-12(-3)=36. The axis of the parabola is the x-axis.

V(0,0),\ F(-3,0),\ \text{directrix }x=3,\ \text{latus rectum}=12

2026 Annual I, Set B, Q.2 (xi)4 marks

A whispering gallery (elliptical) has a ceiling height of 8\,\mathrm{m} at center and width 20\,\mathrm{m}. Where are the foci located?

Solution

Place the centre of the floor at the origin with the floor along the x-axis. The ceiling is the upper half of the ellipse \dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1.
The width is the major axis and the central height is the semi-minor axis.
2a=20\;\Longrightarrow\;a=10,\qquad b=8
For an ellipse c^2=a^2-b^2, where c is the distance from the centre to each focus.
c^2=100-64=36\;\Longrightarrow\;c=6
The foci lie on the major axis (the floor line).
F(\pm6,\,0)

The ceiling is the upper half of an ellipse with a=10, b=8. The foci are on the floor, 6\,\mathrm{m} from the centre.

So each focus is 6\,\mathrm{m} from the centre of the room along the floor, that is 4\,\mathrm{m} from each end wall. A whisper at one focus is reflected by the ceiling to the other focus, 12\,\mathrm{m} away.

\text{Foci at }(\pm6,0):\ 6\,\mathrm{m}\text{ on each side of the centre}

2026 Annual I, Set B, Q.3 OR8 marks

A circle passes through (0,0), (2,0), and (0,2). (a) Find its equation. (b) Find the tangent at (0,2). (c) Find the normal at (0,2).

Solution

(a) Equation of the circle

Let the circle be x^2+y^2+2gx+2fy+c=0 and substitute the three points.
\begin{aligned}(0,0):&\;\;c=0\\(2,0):&\;\;4+4g+c=0\;\Rightarrow\;g=-1\\(0,2):&\;\;4+4f+c=0\;\Rightarrow\;f=-1\end{aligned}
Substitute g=f=-1, c=0.
x^2+y^2-2x-2y=0
Centre (-g,-f)=(1,1) and radius \sqrt{1+1-0}=\sqrt2. Standard form:
(x-1)^2+(y-1)^2=2
Quick reason: the angle at (0,0) between the axes is 90^\circ, so the chord from (2,0) to (0,2) is a diameter and its midpoint (1,1) is the centre.

x^2+y^2-2x-2y=0

(b) Tangent at (0,2)

Use xx_1+yy_1+g(x+x_1)+f(y+y_1)+c=0 with (x_1,y_1)=(0,2).
0\cdot x+2y-(x+0)-(y+2)+0=0\;\Longrightarrow\;-x+y-2=0
So
x-y+2=0\quad(y=x+2)
Check by slopes: radius from (1,1) to (0,2) has slope \tfrac{2-1}{0-1}=-1, so the tangent has slope 1 and passes through (0,2): y=x+2.

x-y+2=0

The normal is perpendicular to the tangent (slope -1) and passes through the centre (1,1) and the point (0,2).
y-2=-1(x-0)
So
x+y-2=0

x+y-2=0

x^2+y^2-2x-2y=0;\ \text{tangent }x-y+2=0;\ \text{normal }x+y-2=0

2026 Annual I, Set D, MCQ 81 mark

For the hyperbola \dfrac{y^2}{36}-\dfrac{x^2}{49}=1, the slopes of the asymptotes are:

A\pm\dfrac76
B\pm\dfrac{36}{49}
C\pm\dfrac{49}{36}
D\pm\dfrac67Answer

Why

Put the right side equal to zero: \dfrac{y^2}{36}=\dfrac{x^2}{49}, so y=\pm\dfrac67x. The slopes are \pm\dfrac67.

For this vertical hyperbola, \dfrac{y^2}{a^2}-\dfrac{x^2}{b^2}=1 with a=6, b=7, the asymptotes are y=\pm\dfrac{a}{b}x, not \pm\dfrac{b}{a}x. Choosing \pm\tfrac76 (option A) is the common mistake.

2026 Annual I, Set D, MCQ 161 mark

The slope of the tangent to the circle x^2+y^2=169 at the point (5,-12) is:

A-\dfrac{12}{5}
B\dfrac{5}{12}Answer
C-\dfrac{5}{12}
D\dfrac{12}{5}

Why

\dfrac{dy}{dx}=-\dfrac{x}{y}=-\dfrac{5}{-12}=\dfrac{5}{12}. Check: the radius to (5,-12) has slope -\tfrac{12}{5}, and \left(-\tfrac{12}{5}\right)\left(\tfrac{5}{12}\right)=-1.

Option C forgets that y_1=-12 is negative.

2026 Annual I, Set D, MCQ 191 mark

The length of the tangent from the point (0,13) to the circle x^2+y^2=144 is:

A\sqrt{25}Answer
B\sqrt5
C\sqrt{169}
D\sqrt{313}

Why

Length =\sqrt{x_1^2+y_1^2-144}=\sqrt{0+169-144}=\sqrt{25}=5 (the 5, 12, 13 right triangle). Option D adds 144 instead of subtracting it.

2026 Annual I, Set D, Q.2 (iv) OR4 marks

Examine whether the line y=-2x+5 cuts, is tangent to, or lies outside the circle x^2+y^2=17.

Solution

The circle has centre (0,0) and radius r=\sqrt{17}\approx4.12.
Line in general form: 2x+y-5=0. Distance of the centre from it:
d=\dfrac{|0+0-5|}{\sqrt{4+1}}=\dfrac{5}{\sqrt5}=\sqrt5\approx2.24
Since d<r, the line cuts the circle at two points.
\sqrt5<\sqrt{17}
Confirm by substitution of y=5-2x.
x^2+(5-2x)^2=17\;\Longrightarrow\;5x^2-20x+8=0,\qquad\Delta=400-160=240>0
Points of intersection.
x=\dfrac{20\pm\sqrt{240}}{10}=2\pm\dfrac{2\sqrt{15}}{5},\qquad y=5-2x=1\mp\dfrac{4\sqrt{15}}{5}
Approximately (0.45,\,4.10) and (3.55,\,-2.10).

\text{Cuts at }\left(2\pm\tfrac{2\sqrt{15}}{5},\ 1\mp\tfrac{4\sqrt{15}}{5}\right)

2026 Annual I, Set D, Q.2 (vii)4 marks

For the parabola x^2=-8y, find the vertex, focus, directrix, and length of the latus rectum.

Solution

Compare with x^2=-4ay, the parabola that opens downwards.
4a=8\;\Longrightarrow\;a=2
Vertex:
V(0,0)
Focus: a units below the vertex on the axis x=0.
F(0,-a)=(0,-2)
Directrix: a units above the vertex.
y=2\quad(y-2=0)
Length of latus rectum.
|4a|=8
Ends of the latus rectum: put y=-2, x^2=16, so (\pm4,-2). The axis is the y-axis.

V(0,0),\ F(0,-2),\ \text{directrix }y=2,\ \text{latus rectum}=8

2026 Annual I, Set D, Q.2 (xi)4 marks

A satellite orbits Earth in an elliptical path with Earth's center at one focus. Major axis length =30{,}000\,\mathrm{km}, minor axis length =20{,}000\,\mathrm{km}. Find distance from Earth's center to ellipse center.

Solution

Semi-axes from the given lengths.
2a=30\,000\;\Longrightarrow\;a=15\,000\,\mathrm{km},\qquad2b=20\,000\;\Longrightarrow\;b=10\,000\,\mathrm{km}
Earth's centre is a focus, so the required distance is c, the distance from the centre of the ellipse to a focus.
c^2=a^2-b^2
Substitute.
c^2=(15\,000)^2-(10\,000)^2=225\times10^{6}-100\times10^{6}=125\times10^{6}
Take the square root.
c=\sqrt{125\times10^{6}}=5000\sqrt5\approx11\,180\,\mathrm{km}
Eccentricity of the orbit, for interest: e=\dfrac{c}{a}=\dfrac{\sqrt5}{3}\approx0.75.

Units: thousands of km. Earth's centre is at the focus, c\approx11\,180\,\mathrm{km} from the centre O.

c=5000\sqrt5\approx11\,180\,\mathrm{km}

2026 Annual I, Set D, Q.3 OR8 marks

A circle has equation: x^2+y^2-6x+4y-12=0. (a) Find the equation of the tangent to the circle at the point P(7,-1). (b) Find the equation of the normal to the circle at the same point P(7,-1).

Solution

Step 1: check the point

Centre and radius: g=-3, f=2, c=-12.
C(3,-2),\qquad r=\sqrt{9+4+12}=5
Substitute P(7,-1).
49+1-42-4-12=-8<0
The value is negative, so P is inside the circle (|CP|=\sqrt{16+1}=\sqrt{17}<5). A tangent at P does not exist; the paper has a misprint. The two single-sign corrections below make P a point of the circle, and both are solved. (With the printed data, the formula xx_1+yy_1+g(x+x_1)+f(y+y_1)+c=0 gives 4x+y-35=0, which is not a tangent: its distance from C is \tfrac{34}{\sqrt{17}}\approx8.2\neq5.)

Version 1: circle x^2+y^2-6x-4y-12=0 (sign of 4y corrected), point P(7,-1)

Check: 49+1-42+4-12=0, so P is on it. Centre C(3,2), r=5.
(a) Tangent: xx_1+yy_1+g(x+x_1)+f(y+y_1)+c=0 with g=-3, f=-2, c=-12.
7x-y-3(x+7)-2(y-1)-12=0\;\Longrightarrow\;4x-3y-31=0
(b) Normal: through P(7,-1) and C(3,2); slope \dfrac{2+1}{3-7}=-\tfrac34.
y+1=-\tfrac34(x-7)\;\Longrightarrow\;3x+4y-17=0
Check: slopes \tfrac43 and -\tfrac34 multiply to -1.

Version 2: printed circle, point P(7,1) (sign of y_1 corrected)

Check: 49+1-42+4-12=0, so (7,1) is on the printed circle with centre C(3,-2).
(a) Tangent with g=-3, f=2, c=-12.
7x+y-3(x+7)+2(y+1)-12=0\;\Longrightarrow\;4x+3y-31=0
(b) Normal through (7,1) and (3,-2); slope \tfrac34.
y-1=\tfrac34(x-7)\;\Longrightarrow\;3x-4y-17=0

P(7,-1) lies inside the printed circle (solid) but on the circle x^2+y^2-6x-4y-12=0 (dashed).

\begin{aligned}\text{V1: }&\text{tangent }4x-3y-31=0\\ &\text{normal }3x+4y-17=0\\ \text{V2: }&\text{tangent }4x+3y-31=0\\ &\text{normal }3x-4y-17=0\end{aligned}

Note. P(7,-1) does not lie on the printed circle (it is inside it). Either the circle should read x^2+y^2-6x-4y-12=0 or the point should be (7,1). The method and marks are the same for both.

2026 Annual I, Old syllabus (Curriculum 2000), MCQ 6, old syllabus1 mark

For parabola y^2=4ax, the focus is at:

A(a,0)Answer
B(-a,0)
C(0,-a)
D(0,a)

Why

y^2=4ax (with a>0) opens to the right, its axis is the x-axis, the vertex is the origin, the focus is (a,0) and the directrix is x=-a. Option D is the focus of x^2=4ay.

2026 Annual I, Old syllabus (Curriculum 2000), MCQ 8, old syllabus1 mark

The equation x^2+y^2=9 represents a:

Aparabola
BcircleAnswer
Chyperbola
Dstraight line

Why

x^2+y^2=9=3^2 is the standard form x^2+y^2=r^2: a circle with centre (0,0) and radius 3. The squared terms have equal coefficients and no xy term.

2026 Annual I, Old syllabus (Curriculum 2000), MCQ 13, old syllabus1 mark

For the ellipse \dfrac{x^2}{16}+\dfrac{y^2}{9}=1, length of the major axis is:

A3
B4
C6
D8Answer

Why

The larger denominator is a^2=16, so a=4 and the major axis has length 2a=8. Option B is only the semi-major axis a; option C is the minor axis 2b.

2026 Annual I, Old syllabus (Curriculum 2000), MCQ 20, old syllabus1 mark

A circle has equation x^2+y^2=25. The point P(3,4) lies:

Ainside the circle
Boutside the circle
Con the circleAnswer
Dat the center

Why

Substitute: 3^2+4^2-25=9+16-25=0. The value is zero, so P lies on the circle (a negative value would mean inside, a positive value outside).

2026 Annual I, Old syllabus (Curriculum 2000), Q.2 (ii) OR, old syllabus4 marks

For the circle x^2+y^2=25, write the equations of the tangent and normal at the point (3,4).

Solution

Check the point: 9+16=25, so (3,4) is on the circle.
The tangent to x^2+y^2=r^2 at (x_1,y_1) is xx_1+yy_1=r^2.
3x+4y=25\quad\text{i.e. }3x+4y-25=0
The normal at a point of a circle passes through the centre (0,0). Slope of the radius =\tfrac43.
y=\tfrac43x\quad\text{i.e. }4x-3y=0
Check: slopes -\tfrac34 and \tfrac43 multiply to -1, so tangent and normal are perpendicular.

\text{Tangent }3x+4y-25=0,\quad\text{normal }4x-3y=0

2026 Annual I, Old syllabus (Curriculum 2000), Q.2 (vi), old syllabus4 marks

A parabola has its focus at F(2,3) and vertex at V(2,1). Write equation of the parabola in standard form.

Solution

V and F have the same x-coordinate, so the axis is the vertical line x=2. The focus is above the vertex, so the parabola opens upwards.
a is the distance from the vertex to the focus.
a=|VF|=3-1=2
Standard form of an upward parabola with vertex (h,k).
(x-h)^2=4a(y-k)
Substitute h=2, k=1, a=2.
(x-2)^2=8(y-1)
Expanded form, if required.
x^2-4x-8y+12=0
The directrix is y=k-a=-1. Check with the definition: for the point (6,3) on the curve, distance to F is 4 and distance to y=-1 is 4.

(x-2)^2=8(y-1)

2026 Annual I, Old syllabus (Curriculum 2000), Q.2 (ix) OR, old syllabus4 marks

Find the equation of an ellipse with center at (2,-3), major axis parallel to the x-axis, length of major axis =10 units, and length of minor axis =6 units.

Solution

Major axis parallel to the x-axis, so a^2 goes under the x term.
\dfrac{(x-h)^2}{a^2}+\dfrac{(y-k)^2}{b^2}=1
From the lengths:
2a=10\Rightarrow a=5,\qquad2b=6\Rightarrow b=3
Centre (h,k)=(2,-3).
\dfrac{(x-2)^2}{25}+\dfrac{(y+3)^2}{9}=1
General form: multiply by 225 and expand.
9(x-2)^2+25(y+3)^2=225\;\Longrightarrow\;9x^2+25y^2-36x+150y+36=0
Its foci: c=\sqrt{25-9}=4, so (2\pm4,-3), that is (6,-3) and (-2,-3); vertices (7,-3) and (-3,-3).

\dfrac{(x-2)^2}{25}+\dfrac{(y+3)^2}{9}=1\ \text{ or }\ 9x^2+25y^2-36x+150y+36=0

2026 Annual I, Old syllabus (Curriculum 2000), Q.4 OR, old syllabus8 marks

Find the points of intersection of the following two conics: \dfrac{x^2}{16}+\dfrac{y^2}{9}=1; \dfrac{x^2}{4}-\dfrac{y^2}{5}=1

Solution

The first curve is an ellipse (a=4, b=3) and the second a hyperbola (a=2, b=\sqrt5), both centred at the origin. Both equations contain only x^2 and y^2, so treat x^2 and y^2 as the unknowns of a pair of linear equations.

Clear the fractions.
\begin{aligned}9x^2+16y^2&=144\quad(1)\\5x^2-4y^2&=20\quad(2)\end{aligned}
Multiply (2) by 4 so the y^2 terms cancel on adding.
20x^2-16y^2=80\quad(3)
Add (1) and (3).
29x^2=224\;\Longrightarrow\;x^2=\dfrac{224}{29}
Substitute in (2).
4y^2=5\cdot\dfrac{224}{29}-20=\dfrac{1120-580}{29}=\dfrac{540}{29}\;\Longrightarrow\;y^2=\dfrac{135}{29}
Take square roots. Both values are positive, so there are four real points, symmetric in both axes.
x=\pm\sqrt{\dfrac{224}{29}}=\pm4\sqrt{\dfrac{14}{29}}\approx\pm2.78,\qquad y=\pm\sqrt{\dfrac{135}{29}}=\pm3\sqrt{\dfrac{15}{29}}\approx\pm2.16
Check with (1): 9(7.724)+16(4.655)=69.52+74.48=144.

\left(\pm4\sqrt{\tfrac{14}{29}},\ \pm3\sqrt{\tfrac{15}{29}}\right)\approx(\pm2.78,\ \pm2.16)\ \text{(four points)}

2026 Annual I, Old syllabus (Curriculum 2000), Q.6 OR, old syllabus8 marks

Prove that an angle in a semi circle is a right angle.

Solution

Statement

If AB is a diameter of a circle and P is any other point on the circle, then \angle APB=90^\circ.

Figure and set-up

Take the centre at the origin and the diameter along the x-axis.

Choose axes so that the centre is the origin O(0,0) and the diameter AB lies on the x-axis. If the radius is r, the circle is x^2+y^2=r^2 and the ends of the diameter are A(-r,0) and B(r,0). Let P(x_1,y_1) be any point of the circle other than A and B, so y_1\neq0 and

x_1^2+y_1^2=r^2\quad(1)

Proof

Slope of AP, joining A(-r,0) to P(x_1,y_1).
m_1=\dfrac{y_1-0}{x_1+r}=\dfrac{y_1}{x_1+r}
Slope of BP, joining B(r,0) to P(x_1,y_1).
m_2=\dfrac{y_1}{x_1-r}
Multiply the slopes. The denominator is a difference of squares.
m_1m_2=\dfrac{y_1^2}{x_1^2-r^2}
From (1), x_1^2-r^2=-y_1^2. Substitute.
m_1m_2=\dfrac{y_1^2}{-y_1^2}=-1
Two lines whose slopes multiply to -1 are perpendicular, so AP\perp BP.
\angle APB=90^\circ

Alternative proof with vectors

Position vectors relative to P.
\overrightarrow{PA}=(-r-x_1,\,-y_1),\qquad\overrightarrow{PB}=(r-x_1,\,-y_1)
Dot product, using (1).
\overrightarrow{PA}\cdot\overrightarrow{PB}=(-r-x_1)(r-x_1)+y_1^2=x_1^2-r^2+y_1^2=0
A zero dot product of non-zero vectors means they are perpendicular, so \angle APB=90^\circ.

Converse (useful fact)

If \angle APB=90^\circ for a moving point P, then P lies on the circle with diameter AB. This is why the circle through (0,0), (2,0) and (0,2) has the chord from (2,0) to (0,2) as a diameter (2026 Set B, Q.3 OR).

m_{AP}\cdot m_{BP}=-1\ \Rightarrow\ \angle APB=90^\circ

2026 Annual I, Old syllabus (Curriculum 2006), MCQ 12, old syllabus1 mark

For what value of c, line y=2x+c is tangent to the circle x^2+y^2=9?

A\pm3
B\pm\sqrt5
C\pm5
D\pm3\sqrt5Answer

Why

Condition of tangency for y=mx+c and x^2+y^2=r^2: c^2=r^2\left(1+m^2\right)=9(1+4)=45, so c=\pm\sqrt{45}=\pm3\sqrt5.

2026 Annual I, Old syllabus (Curriculum 2006), MCQ 13, old syllabus1 mark

The centre and radius of the circle x^2+y^2+4x-8y+11=0 is:

A(-2,4);\ 3Answer
B(2,-4);\ 3
C(-2,4);\ 9
D(2,-4);\ 9

Why

2g=4, 2f=-8, so g=2, f=-4 and the centre is (-g,-f)=(-2,4). Radius =\sqrt{g^2+f^2-c}=\sqrt{4+16-11}=\sqrt9=3. Option C gives r^2 instead of r.

2026 Annual I, Old syllabus (Curriculum 2006), MCQ 14, old syllabus1 mark

The equation of tangent to a parabola y^2=4x at point P(1,2) is:

Ay=2x+1
By=2x
Cy=x-1
Dy=x+1Answer

Why

Here 4a=4, a=1. The tangent to y^2=4ax at (x_1,y_1) is yy_1=2a(x+x_1): 2y=2(x+1), so y=x+1. Check: it passes through (1,2) and y=x+1 in y^2=4x gives (x-1)^2=0, a double root.

2026 Annual I, Old syllabus (Curriculum 2006), MCQ 15, old syllabus1 mark

What is the eccentricity of an ellipse when it becomes a circle?

Ae=1
Be<1
Ce>1
De=0Answer

Why

For an ellipse e=\dfrac{c}{a}=\dfrac{\sqrt{a^2-b^2}}{a}. When b=a the ellipse is a circle, c=0 and both foci coincide with the centre, so e=0.

2026 Annual I, Old syllabus (Curriculum 2006), Q.2 (v) OR, old syllabus4 marks

Find equation of tangent to the circle x^2+y^2-6x-8y+9=0 at the point (5,3).

Solution

Checking the point

Substitute (5,3) in the left side.
25+9-30-24+9=-11<0
The value is negative, so (5,3) lies inside the circle (centre (3,4), r=\sqrt{9+16-9}=4). No tangent can be drawn from a point inside a circle, so the question as printed has no answer.
Applying xx_1+yy_1+g(x+x_1)+f(y+y_1)+c=0 mechanically gives 2x-y-18=0. Its distance from the centre is \tfrac{|6-4-18|}{\sqrt5}\approx7.2\neq4, so it is not a tangent and would not earn the marks as a correct answer.

Corrected question: circle x^2+y^2-6x-8y+20=0

With constant term 20 the point lies on the circle: 25+9-30-24+20=0. Centre (3,4), r=\sqrt{9+16-20}=\sqrt5.
Tangent at (x_1,y_1)=(5,3) with g=-3, f=-4, c=20.
5x+3y-3(x+5)-4(y+3)+20=0\;\Longrightarrow\;2x-y-7=0
Check: slope of the radius from (3,4) to (5,3) is -\tfrac12; slope of the tangent is 2; product -1.

(5,3) is inside the printed circle (dashed) but on the corrected circle (solid).

\text{As printed: no tangent (point inside). Corrected: }2x-y-7=0

Note. The point (5,3) does not lie on the printed circle. Most likely the constant term should be 20 instead of 9; the tangent is then 2x-y-7=0.

2026 Annual I, Old syllabus (Curriculum 2006), Q.2 (ix) OR, old syllabus4 marks

Find equation of a parabola with focus (0,1) and directrix y+1=0.

Solution

By definition, a point P(x,y) is on the parabola when its distance from the focus equals its distance from the directrix.
|PF|=|PM|
Distance to F(0,1) and distance to the line y=-1.
\sqrt{(x-0)^2+(y-1)^2}=|y+1|
Square both sides.
x^2+y^2-2y+1=y^2+2y+1
Simplify.
x^2=4y
This is x^2=4ay with a=1: vertex (0,0), axis the y-axis, opening upwards, latus rectum 4.

x^2=4y

2026 Annual I, Old syllabus (Curriculum 2006), Q.2 (x) OR, old syllabus4 marks

Find center and radius of the circle 2x^2+2y^2-\tfrac52x+\tfrac32y-\tfrac74=0.

Solution

Divide by 2 so that the coefficients of x^2 and y^2 are 1.
x^2+y^2-\tfrac54x+\tfrac34y-\tfrac78=0
Compare with x^2+y^2+2gx+2fy+c=0.
2g=-\tfrac54\Rightarrow g=-\tfrac58,\qquad2f=\tfrac34\Rightarrow f=\tfrac38,\qquad c=-\tfrac78
Centre (-g,-f).
\left(\tfrac58,-\tfrac38\right)
Radius.
r=\sqrt{g^2+f^2-c}=\sqrt{\tfrac{25}{64}+\tfrac{9}{64}+\tfrac{56}{64}}=\sqrt{\tfrac{90}{64}}=\dfrac{3\sqrt{10}}{8}\approx1.19

\text{Centre }\left(\tfrac58,-\tfrac38\right),\quad r=\dfrac{3\sqrt{10}}{8}

2026 Annual I, Old syllabus (Curriculum 2006), Q.2 (xi) OR, old syllabus4 marks

Find equation of tangent to the hyperbola \tfrac23x^2-\tfrac35y^2=1 at point \left(\tfrac{3\sqrt6}{2},\tfrac53\right).

Solution

Checking the point

Substitute the printed point.
\tfrac23\cdot\tfrac{54}{4}-\tfrac35\cdot\tfrac{25}{9}=9-\tfrac53=\tfrac{22}{3}\neq1
So \left(\tfrac{3\sqrt6}{2},\tfrac53\right) is not on the hyperbola, and a tangent at it does not exist. The point of the hyperbola with ordinate \tfrac53 is found from
\tfrac23x^2=1+\tfrac35\cdot\tfrac{25}{9}=\tfrac83\;\Longrightarrow\;x^2=4\;\Longrightarrow\;x=\pm2
We take the intended point as (2,\tfrac53).

Tangent at (2,\tfrac53)

For Ax^2-By^2=1 the tangent at (x_1,y_1) is obtained by replacing x^2 by xx_1 and y^2 by yy_1: Axx_1-Byy_1=1. (It also follows by implicit differentiation: \tfrac43x-\tfrac65yy'=0, so y'=\tfrac{10x}{9y}=\tfrac{20}{15}=\tfrac43 at the point.)
Substitute x_1=2, y_1=\tfrac53.
\tfrac23(2)x-\tfrac35\left(\tfrac53\right)y=1\;\Longrightarrow\;\tfrac43x-y=1
Multiply by 3.
4x-3y-3=0
Check: 4(2)-3\left(\tfrac53\right)-3=8-5-3=0, and its slope \tfrac43 agrees with y'.

4x-3y-3=0\quad\text{(at the point }(2,\tfrac53)\text{)}

Note. The printed point \left(\tfrac{3\sqrt6}{2},\tfrac53\right) does not satisfy the equation. The point on the curve with y=\tfrac53 in the first quadrant is (2,\tfrac53), used here.

2026 Annual I, Old syllabus (Curriculum 2006), Q.2 (xii), old syllabus4 marks

Find equation of an ellipse with foci (\pm2,0) and vertices (\pm3,0).

Solution

Foci and vertices lie on the x-axis, symmetric about the origin, so the centre is (0,0) and the major axis is along the x-axis.
\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1
Vertices (\pm a,0) and foci (\pm c,0) give
a=3,\qquad c=2
For an ellipse b^2=a^2-c^2.
b^2=9-4=5
Equation.
\dfrac{x^2}{9}+\dfrac{y^2}{5}=1\quad\text{or}\quad5x^2+9y^2=45
Eccentricity e=\tfrac{c}{a}=\tfrac23<1, as it must be for an ellipse.

\dfrac{x^2}{9}+\dfrac{y^2}{5}=1

2026 Annual I, Old syllabus (Curriculum 2006), Q.4 OR, old syllabus8 marks

Find equation of a circle passing through points A(1,2), B(4,3) and C(2,5).

Solution

Let the circle be x^2+y^2+2gx+2fy+c=0. Substitute each point.
\begin{aligned}A(1,2):&\;\;5+2g+4f+c=0\quad(1)\\B(4,3):&\;\;25+8g+6f+c=0\quad(2)\\C(2,5):&\;\;29+4g+10f+c=0\quad(3)\end{aligned}
Subtract (1) from (2) and (1) from (3) to remove c.
\begin{aligned}20+6g+2f&=0\;\Rightarrow\;3g+f=-10\quad(4)\\24+2g+6f&=0\;\Rightarrow\;g+3f=-12\quad(5)\end{aligned}
From (4), f=-10-3g. Substitute in (5).
g-30-9g=-12\;\Longrightarrow\;-8g=18\;\Longrightarrow\;g=-\tfrac94
Then
f=-10+\tfrac{27}{4}=-\tfrac{13}{4}
From (1).
c=-5-2g-4f=-5+\tfrac92+13=\tfrac{25}{2}
Substitute and multiply by 2.
x^2+y^2-\tfrac92x-\tfrac{13}{2}y+\tfrac{25}{2}=0\;\Longrightarrow\;2x^2+2y^2-9x-13y+25=0
Centre and radius.
\left(\tfrac94,\tfrac{13}{4}\right),\qquad r=\sqrt{\tfrac{81}{16}+\tfrac{169}{16}-\tfrac{200}{16}}=\sqrt{\tfrac{50}{16}}=\dfrac{5\sqrt2}{4}\approx1.77
Check with B: 2(16)+2(9)-36-39+25=0. With C: 8+50-18-65+25=0.

2x^2+2y^2-9x-13y+25=0

Exam focus

Derive the standard equation of the parabola y^2=4ax and find its elements (vertex, focus, directrix, latus rectum) for a given parabola. (long)
Both 2026 new sets asked for the elements of a parabola (y^2=-12x, x^2=-8y) in Q.2 (vii); the derivation is a standard long question.
Go to answer
Find the equation of the tangent and normal to a circle at a given point (check first that the point lies on the circle). (problem)
An 8-mark Section C question in both new sets (Q.3 OR) and in the old papers; Set D and the old 2006 paper printed points not on the circle.
Go to answer
Decide whether a line cuts, touches or misses a circle, and find the points of intersection. (problem)
Set B and Set D Q.2 (iv) OR (4 marks); also Exercise 7.2 Q1 and Q2.
Go to answer
Find the equation of a circle through three points, or through two points with its centre on a given line. (problem)
Set B Q.3 OR and old 2006 Q.4 OR (8 marks); Exercise 7.1 Q3 to Q7.
Go to answer
Find the length of the tangent from an external point and the slope of a tangent to x^2+y^2=r^2. (mcq)
Four MCQs in the two 2026 new sets used these formulas.
Go to answer
Find the foci of an ellipse from its axes in an application (whispering gallery, satellite orbit). (problem)
Q.2 (xi) in both 2026 new sets: c^2=a^2-b^2 in a real setting.
Go to answer
Derive the standard equation of the ellipse and of the hyperbola and list their elements, including the asymptotes of a hyperbola. (long)
Asymptote slopes were MCQs in both new sets; the derivations are standard long questions.
Go to answer
Write the conditions of tangency c^2=r^2(1+m^2), c=a/m, c^2=a^2m^2\pm b^2 and use them to find tangents in slope form. (short)
Old 2006 MCQ 12 and Exercise 7.2 Q3; one idea covers all four conics.
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Convert a second degree equation to the standard form of an ellipse or hyperbola, find its elements and draw the graph. (problem)
A common Section B question (old 2000 Q.2 (ix) OR asked the reverse: write the ellipse from its centre and axes).
Go to answer
Prove that the tangent at any point of a circle is perpendicular to the radius, and that an angle in a semicircle is a right angle. (long)
Exercise 7.2 Q8 and old 2000 Q.6 OR (8 marks): short coordinate proofs using slopes.
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