Mathematics

Session

HyperbolaSession - 2

Session Objectives

Session Objectives

1. Equation of chord joining two points on the hyperbola

2. Equation of chord whose mid-point is given

3. Equation of pair of tangents from an external point

4. Equation of chord of contact

5. Asymptotes of hyperbola

6. Rectangular hyperbola

7. Equation of rectangular hyperbola referred to its asymptotes as the axes of coordinates

8. Director circle

Equation of the Chord Joining Two

Points on the Hyperbola 12 2

2 2

x y

a b

The equation of the chord joining twopoints

on the hyperbola is

1 1 2 2P asec , b tan and Q asec , b tan

2 2

2 2

x y1

a b

2 1

1 12 1

b tan b tany b tan x asec

asec asecwhich reduces to

1 2 1 2 1 2x ycos sin cos

a 2 b 2 2

Equation of Chord whose Mid-Point is Given

Equation of chord of the hyperbola

whose middle point is

is given by

2 2

2 2

x y1

a b 1 1x , y

2 2

1 1 1 12 2 2 2

xx yy x y1 1

a b a b

1 12 2

xx yyi.e. , where T 1

a b1T = S

2 21 1

1 2 2

x yS 1

a b

Equation of Pair of Tangents Fromand External Point

Equation of pair of tangents from the

point to the hyperbola

is

1 1x , y 2 2

2 2

x y1

a b

22 22 21 1 1 1

2 2 2 2 2 2

x y xx yyx y1 1 1

a b a b a b

i.e. , where 21SS T

2 22 2

1 1 1 112 2 2 2 2 2

xx yy x yx yT 1, S 1, S 1

a b a b a b

Equation Chord of Contact

Equation of chord of contact of point with respect to the hyperbola

is , i.e. T = 0

1 1x , y

2 2

2 2

x y1

a b 1 1

2 2

xx yy1 0

a b

Asymptotes of Hyperbola

An asymptote is a straight line, whichmeets the conic in two points both ofwhich are situated at an infinite distance,but which is itself not altogether(entirely) at infinity.

To Find the Equation of the Asymptotes of the Hyperbola

2 2

2 2

x y= 1

a b

Let the straight line

y = mx + c ... (i)

meet the given hyperbola in points,whose abscissae are given by theequation

22

2 2

mx cx1

a b

or ... (ii) 2 2 2 2 2 2 2 2x b a m 2a mcx a c b 0

To Find the Equation of the Asymptotes of the Hyperbola

2 2

2 2

x y= 1

a b

If the straight line (i) be an asymptote,both roots of equation (ii) must be infinite.

Hence, the coefficients of x2 and x in theequation (ii) must be zero.

We have 2 2 2 2b a m 0 and a mc 0

Hence, and c = 0b

ma

To Find the Equation of the Asymptotes of the Hyperbola

2 2

2 2

x y= 1

a b

Substituting the values of m

and c in

y = mx + c, we get

b

y xa

, i.e.

x y x y

0 and 0a b a b

The combined equation of the

asymptotes is2 2

2 2

x y0

a b

X

Y´

X´

Y

O xa

2

2— – — = 1y

b

2

2

Asymptotes

xa

2

2— – — = – 1y

b

2

2

Asymptotes

Points to Remember

i. A hyperbola and its conjugate hyperbolahave the same asymptotes.

ii. The equation of the pair of asymptotesdiffer the hyperbola and the conjugatehyperbola by the same constant only, i.e

Hyperbola – Asymptotes = Asymptotes – Conjugate Hyperbola

2 2 2 2 2 2 2 2

2 2 2 2 2 2 2 2

x y x y x y x y1 1

a b a b a b a b

iii. The asymptotes pass through the centre of hyperbola.

iv. The bisectors of the angles between the asymptotes of the

hyperbola are the coordinate axes (or axes of the

hyperbola).

2 2

2 2

x y1

a b

Points to Remember

v. As we know that combined equation of

asymptotes is and equation

of hyperbola is

2 2

2 2

x y0

a b

2 2

2 2

x y1.

a b

Equation of pair of asymptotes and equation of hyperbola differ by a constant only. (Important)

Rectangular Hyperbola or Equilateral Hyperbola

A hyperbola whose asymptotes are atright angles to each other is called arectangular hyperbola.

The equations of asymptotes of the hyperbola

are given by 2 2

2 2

x y1

a b

by x.

a

The angle between two asymptotes is given by

2 2 2

2

b b 2b2aba a atan

b b b a b1 1a a a

Rectangular Hyperbola or Equilateral Hyperbola

If the asymptotes are at right angles,

then 2 2tan a b 0

2

a b 2a 2b

Thus, the transverse and conjugate axesof a rectangular hyperbola are equal.

2 2 2The equation of rectangular hyperbola is x y a .

The equation of the asymptotes of the rectangular

hyperbola are y x.

Cor: Eccentricity of rectangular hyperbola is 2.

Equation of the Rectangular Hyperbola Referred to its Asymptotes as the Axes of Coordinates

Referred to the transverse and conjugateaxes as the axes of coordinates, theequation of the rectangular hyperbola is

2 2 2x y a ... (i)

The equation of asymptotes of the hyperbola (i) is

Each of these two asymptotes is inclined at an angle

of with the transverse axis.

y x.

4

Equation of the Rectangular Hyperbola Referred to its Asymptotes as the Axes of Coordinates

Now rotating the axes through an angle in

clockwise direction, keeping the origin fixed,then the axes coincide with the asymptotes

of the hyperbola and

4

X Yx Xcos Y sin

4 4 2

Y Xand y Xsin Ycos

4 4 2

Putting the values of x and y in (i), we get

2 22X Y Y X

a2 2

Equation of the Rectangular Hyperbola Referred to its Asymptotes as the Axes of Coordinates

2a

XY2

2

2 2 aXY c where c

2

X

Y´

X´

Y

O

This is the transformed equationof rectangular hyperbola (i).

Thus, equation of rectangularhyperbola when its asymptotestaken as coordinate axes is

2xy c .

Cor: If equation of a rectangularhyperbola be thenequation of its conjugatehyperbola will be

2xy c ,

2xy c .

Parametric Form of RectangularHyperbola xy = c2

If ‘t’ is non-zero variable, the coordinates ofany point on the rectangular hyperbola

xy = c2 can be written

cct, .

t

The point is also called point ‘t’.

cct,

t

Equation of Chord Joining Points ‘t1’ and ‘t2’

The equation of the chord joining two

points and of hyperbola

xy = c2 is

11

cct ,

t

22

cct ,

t

2 11

1 2 1

c ct tc

y x ctt ct ct

1 2 1 2x t t y c t t

This is the required equation of chord.

Equation of Tangent in Different Forms

(i) Equation of tangent in point formof the hyperbola xy = c2

21 1xy x y 2c

(ii) Equation of tangent in parametric form

xyt 2c

t

Equation Normal in Different Forms

(i) Equation of normal in point form

2 21 1 1 1xx yy x y

Equation Normal in Parametric Forms

(i) Equation of normal in parametricform

3 4xt yt ct c 0

Note: The equation of normal at is a

fourth degree equation in t. Therefore, in general four normal can be drawn from a point to the hyperbola xy = c2.

cct,

t

Point of Intersection of Tangents at ‘t1’ and ‘t2’ to the Hyperbola xy = c2

The equations to the tangents at thepoints ‘t1’ and ‘t2’ are

1 21 2

x xyt 2c and y t 2c

t t

By solving these equations, we get point ofintersection of tangents.

Coordinates of point of intersection of tangents at ‘t1’ and ‘t2’ is .

1 2

1 2 1 2

2ct t 2c,

t t t t

Director Circle

The locus of intersection of tangentswhich are at right angles is calleddirector circle of Hyperbola.

To find the locus of the point of intersectionof tangents which meet at right angles.

2 2

2 2

x yLet equation of hyperbola is 1 ...(i)

a b

2 2 2Any tangent to the hyperbola is y mx a m b ...(ii)

22 21 1

and its perpendicular tangent is y x a b ... (iii)m m

Director Circle

Let (h, k) be their point of intersection.We have

2 2 2k mh a m b ... (iv)

2 2 2and mk h a b m ... (v)

[By putting the value of (h, k) in equations (iii) and (iv)]

If between (iv) and (v), we eliminate m, we shall have a relation between h and k, i.e. locus of (h, k).

Squaring and adding these equations, we get

2 2 2 2 2 2 2 2k mh mk h a m b a b m

Director Circle

2 2 2 2 2 2k h 1 m a b 1 m

2 2 2 2h k a b

Locus of (h, k) is 2 2 2 2x y a b

This is the equation of director circle, whose centre

is origin and radius is 2 2a b .

Class Test

Class Exercise - 1

If the chord through the points

on the hyperbola passes

through a focus, prove that

asec , b tan and a sec , b tan

2 2

2 2

x y1

a b

1 etan tan .

2 2 1 e

Solution

The equation of the chord joining is

x ycos sin cos

a 2 b 2 2

If it passes through the focus (ae, 0), then

ecos cos2 2

cos12e

cos2

By componendo and dividendo,

cos cos1 e2 21 e

cos cos2 2

1 e

cot cot2 2 1 e

1 e

tan tan2 2 1 e [Proved]

Class Exercise - 2

Chords of the hyperbola touch the parabola Prove thatthe locus of their middle points is thecurve

2 2 2x y a2y 4ax.

2 3y x a x .

Solution

If (h, k) be the mid-point of the chord,then the equation of the chord is T= S1,

2 2 2 2i.e. xh yk a h k a

2 2i.e. xh yk h k

2 2k hhy x = mx + c [Say]

k k

If it touches the parabola y2 = 4ax, then

a

cm

[Condition for tangency for any line y = mx + c to the parabola]

Solution contd..

2 2 2ak h k h

Locus of (h, k) is 2 2 2ay x y x

2 3 i.e. y x a x .

2 2k h akk h

Class Exercise - 3

Find the point of intersection oftangents drawn to the hyperbola

at the points where it

is intersected by the linelx + my + n = 0.

2 2

2 2

x y1

a b

Solution

Let (h, k) be the required point.

Equation of chord of contact drawnfrom (h, k) to the hyperbola is

T = 0

2 2

xh yki.e. 1 0 ... (i)

a b

The given line islx + my + n = 0 ... (ii)

Equations (i) and (ii) represent same line

2 2

h k 1na l b m

Solution contd..

2a l

hn

2b m

kn

Coordinates of the required point

2 2a l b m,

m n

Class Exercise - 4

Prove that the product of theperpendiculars from any point

on the hyperbola to its

asymptotes is equal to

2 2

2 2

x y1

a b

2 2

2 2

a b.

a b

Solution

Let be any point on

the hyperbola

a sec , b tan

2 2

2 2

x y1.

a b

The equation of the asymptotes of the

given hyperbola are x y x y

0 and 0a b a b

1p Length of perpendicular from

x y

a sec , b tan on 0a b

2 2

sec tan

1 1

a b

Solution contd…

2p Length of perpendicular from

a sec , b tan x y

on 0a b

2 2

sec tan

1 1

a b

2 2 2 2

1 2 2 2

2 2

sec tan a bp p Pr oved

1 1 a ba b

Class Exercise - 5

The asymptotes of a hyperbola areparallel to lines 2x + 3y = 0 and3x + 2y = 0. The hyperbola hasits centre at (1, 2) and it passesthrough (5, 3). Find its equation.

Solution

Asymptotes are parallel to lines2x + 3y = 0 and 3x + 2y = 0

Equations of asymptotes are

2x + 3y + k1 = 0 and 3x + 2y + k2 = 0

As we know that asymptotes passes through the centre of the hyperbola.Here centre of hyperbola is (1, 2).

1 22 6 k 0 and 3 4 k 0

1 2k 8, k 7

Solution contd..

The equations of asymptotes are

2x + 3y – 8 = 0 and 3x + 2y – 7 = 0

Equation of hyperbola is

(2x + 3y – 8) (3x + 2y – 7) + c = 0

It passes through (5, 3).

10 9 8 15 6 7 c 0 c 154

Equation of hyperbola is (2x + 3y – 8)(3x + 2y – 7) – 154 = 0

i.e. 6x2 + 13xy + 6y2 – 38x – 37y – 98 = 0

Class Exercise - 6

The chord PP´ of a rectangularhyperbola meets asymptotes inQ and Q´. Show QP = P´Q´.

Solution

Let equation of rectangular hyperbolais xy = c2.

1

1

c cCoordinates of P ct, and P´ ct ,

t t

Equation of chord PP´ is 1 1x y tt c t t 0

It meets asymptotes, i.e. axes at Q and Q´respectively.

1Putting y = 0, Q c t t , 0

1

1 1and putting x = 0, Q´ 0, c

t t

Solution contd..

22

1

cNow PQ ct ct ct

t

21 2

1c t

t

2

2 21

1 1

c c cP´Q´ c t

t t t 21 2

1c t

t

PQ P´Q´ [Proved]

Class Exercise - 7

The normal at the three points P, Q, Ron a rectangular hyperbola, intersectat a point S on the curve. Prove thatcentre of the hyperbola is the centroidof PQR.

Solution

Let equation of the rectangular hyperbolais xy = c2.

Let ‘t’ be the parameter of any of points P, Q, R so that normal is ... (i)

It passes through a point S on the hyperbola.

Let coordinates of point

cS ct́ ,

t́

3 4cFrom i ct́ t .t ct c 0

t́

3 3tt́ t 1 t t́ 1 0

t́

Solution contd..

3 tt́ t 1 1 0

t́

3t t́ 1 0 (Remember this result)

This is a cubic equation in t, and gives us theparameters of the three points P, Q, R, say 1 2 3t , t , t .

2

1 2 3t t t 0 Coeff of t 0

1 2

1 2 3 1 2 3

t t1 1 1and 0 Coeff of t 0

t t t t t t

If (h, k) is the centroid of ,PQR

1 2 3c t t th 0

3

Solution contd..

1 2 3

1 1 1c

t t tk 0

3

Hence, centroid is (0, 0) which is centre of the hyperbola.

Class Exercise - 8

A rectangular hyperbola whose centreis C is cut by any circle of radius r infour points P, Q, R and S. Prove that

2 2 2 2 2CP CQ CR CS 4r .

Solution

Let the equation of rectangular hyperbola is xy = k2 ... (i)

and equation of circle is

2 2x y 2gx 2fy c 0 ... (ii)

where ... (iii) 2 2 2g f c r

Eliminating y between equations (i) and (ii), we get

4 2

22

k kx 2gx 2f c 0

xx

4 3 2 2 4x 2gx cx 2fk x k 0

Solution contd..

This is biquadratic equation in x, whichgives us the abscissae of the four pointsof intersection.

Let x1, x2, x3, x4 are the roots of the equation.

1 2 3 4 1 2x x x x 2g, x x c

22 2 2 21 2 3 4 1 2 3 4 1 2x x x x x x x x 2 x x

24g 2c

Similarly, eliminating x from (i) and (ii), we get

2 2 2 2 21 2 3 4y y y y 4f 2c

Solution contd..

2 2 2 2CP CQ CR CS

2 2 2 2 2 2 2 21 1 2 2 3 3 4 4x y x y x y x y

2 21 1x y

2 24 g f c

24r [Pr oved]

Class Exercise - 9

Prove that the locus of the mid-pointsof the chords of the hyperbola which pass through a fixed

point is a hyperbola whose

centre is

2 2

2 2

x y1

a b

,

, .2 2

Solution

Let (h, k) be the coordinates of mid-pointof the chord.

Equation of chord is T = S1

2 2

2 2 2 2

hx ky h k1 1

a b a b

2 2

2 2 2 2

hx ky h k

a b a b

It passes through a fixed point . ,

2 2

2 2 2 2

h k h k

a b a b

Solution contd..

Locus of (h, k) is

2 2

2 2

x x y y0

a b

2 2

2 22

2 2 4 4

x y12 2

c say4a b a b

This is an equation of hyperbola whose centre is

, [Pr oved]2 2

Class Exercise - 10

From a point, tangents to therectangular hyperbola are drawn and they intersect eachother at an angle of 45o. Prove thatthe locus of the point is the curve

2 2 2x y a

22 2 2 2 2 4x y 4a x y 4a .

Solution

Here equation of rectangular hyperbola is

2 2 2x y a ... (i)

Equation of any tangent to this hyperbola is

2 2 2y mx a m a

If it is passes through (h, k), then 2 2 2 2k mh a m a

2 2 2 2 2m h a 2mkh k a 0

Let m1, m2 be the roots of the above equation, which gives the slopes of two tangents passing through (h, k).

Solution contd..

2 2

1 2 1 22 2 2 2

2kh k am m and m m

h a h a

Given that angle between two tangents are 45o.

1 2

1 2

m mtan45

1 m m 2 2

1 2 1 2m m 1 m m

2 2

1 2 1 2 1 2m m 4m m 1 m m

22 22 2 2 2

2 2 22 22 2

4 k a4h k k a1

h ah ah a

22 2 2 2 2 2 2 24 h k h a k a h k

Locus of (h, k) is 22 2 2 2 2 4x y 4a x y 4a [Pr oved]

Thank you