Mathematics

Session

HyperbolaSession - 1

Introduction

Z

M P

S (tocus)

Z´

Dire

ctrix

If S is the focus, ZZ´ is the directrix and P is any point on the hyperbola, then by definition

SP e SP ePMPM

Question

Illustrative ProblemFind the equation of hyperbola whose focus is (1, 1), directrix is 3x + 4y + 8 = 0 and eccentricity is 2.

Solution :Let S(1, 2) be the focus and P(x, y) be any point on the hyperbola.

SP ePM

where PM = perpendicular distance from P to directrix 3x + 4y + 8 = 0

2 2

2 2

3x 4y 8x 1 y 1 23 4

Solution Cont.

22 2 4 3x 4y 8

x 1 y 125

2 2

2 2

25 x 2x 1 y 2y 1

4 9x 16y 64 24xy 64y 48x

2 211x 96xy 39y 242x 306y 206 0

Ans.

Equation of The Hyperbola in Standard Form

X

Y´

X´

Y

O

Z´ Z

A(a, 0)

S(ae, 0)

NK´ KA´(– a, 0)

S´(– ae, 0)

MM ´P (x, y)

2 2

2 2 22 2

x y 1, where b a e 1 .a b

Definition of Special Points Lines of the Equation of Hyperbola

X

Y´

X´

Y

O

Z´ Z

A(a, 0)

S(ae, 0)

NK´ KA´(– a, 0)

S´(– ae, 0)

MM´P

T

T´L´

L

P´

B

B´ x = –ae

(i) Vertices (ii) Transverse and Conjugate Axes

(iii) Foci : As we have discussed earlier S(ae, 0) and S´(–ae, 0) are the foci of the hyperbola.

Definition of Special Points Lines of the Equation of Hyperbola(iv) Directrices :The lines zk and z´k´ are two directrices of the hyperbola and

their equations are respectively.

a ax and x ,e e

(v) Centre : The middle point O of AA´ bisects every chord of the hyperbola passing through it and is called the centre of the hyperbola.

(vi) EccentricityFor the hyperbola we have

2 2

2 2x y

1,a b

2 2 2b a e 1 .

2 2 22

2 2a b be e 1

a a

Definition of Special Points Lines of the Equation of Hyperbola(vii) Ordinate and Double ordinate

(viii) Latus rectum

(ix) Focal Distance of a Point

SP = ex – a S´P = ex + a

S´P SP ex a ex a 2a Transverse Axis.

“A hyperbola is the locus of a point which moves in such a way that the difference of its distances from two fixed points (foci) is always constant.”

Conjugate Hyperbola

X

Y´

X´

Y

Z´

ZK

K´

B

B´

O

The conjugate hyperbola of the hyperbola

2 2 2 2

2 2 2 2x y x y1 is 1a b a b

The eccentricity of the conjugate hyperbola is 2 2 2a b e 1 .

Important TermsHyperbola Conjugate Hyperbola

(I) Co-ordinates of the centre (0, 0) (0, 0)

(ii) Co-ordinates of the vertices (a, 0) and (–a, 0) (0, b) and (0, –b)(iii) Co-ordinates of foci (ae, 0) and (–ae, 0) (0, be) and (0, –be)

(iv) Length of the transverse axis 2a 2b(v) Length of conjugate axis 2b 2a

(vi) Equation of the directrices

(vii) Eccentricity

(viii) Length of the latus-rectum

(ix) Equation of the transverse axis y = 0 x = 0(x) Equation of the conjugate axis x = 0 y = 0

2 2

2 2x y– 1a b

2 2

2 2x y

– 1a b

axe

bye

2 2 2b a e – 1 2 2 2a b e – 1

22ba

22ab

Auxiliary Circle and Eccentric AngleParametric Coordinate of Hyperbola

The circle drawn on transverse axis of the hyperbola as diameter is called an auxiliary circle of the hyperbola.

If is the equation

Of hyperbola, then its auxiliary circle is x2 + y2 = a2

2 2

2 2x y 1a b

= eccentric angle

x asec , y b tanare known as parametric equation of hyperbola.

Position of Point with respect to Hyperbola

insideinside Outside

2 21 1

1 12 2x yIf 1 0 Point x , y lies inside.a b

2 21 1

1 12 2x yIf 1 0 Point x , y lies on.a b

2 21 1

1 12 2x yIf 1 0 Point x , y lies outside the hyperbola.a b

Intersection of a Line and a Hyperbola

Let the equation of line is y = mx + c

and equation of hyperbola is 2 2

2 2x y 1a b

Point of intersection of line and hyperbola could be found out by solving the above two equations simultaneously.

Intersection of a Line and a Hyperbola

2 2 2 2 2 2 2 2x a m b 2a mcx a c b 0

This is a quadratic equation in x and therefore gives two values of x which may be real and distinct, coincident or imaginary.

22

2 2

mx cx 1 0a b

[Putting the value of y in the equation of Hyperbola]

Condition for Tangency and Equation of Tangent in Slope Form and Point of contact

Given hyperbola is

and given line is y = mx + c

2 2

2 2x y 1a b

2 2 2 2c a m b

2 2 2or c a m b ...............(ii)

This is the required condition for tangency.

Equation of Tangent in Slope FormSubstituting the value of c in the equation y = mx + c, we get equation of tangent in slope form.

Equation of tangent

2 2 2y mx a m b ..................(iii)

Point of Contact

2 2

2 2 2 2 2 2

a m b,a m b a m b

Equation of Tangent and Normal in Point FormEquation of tangent at any point (x1, y1) of the hyperbola is

1 12 2

xx yy 1a b

Equation of Normal at any point (x1, y1) of the hyperbola is

2 21 1

1 1

b y y a x xy x

Equation of Tangent and Normal in Parametric Form

Equation of tangent at

is (asec ,b tan )

x sec y tan 1a b

Equation of normal in parametric form is

2 22 2a x b y a b

asec b tan

2 2axcos bycot a b

Class Test

Class Exercise - 1Find the equation to the hyperbola for which eccentricity is 2, one of the focus is (2, 2) and corresponding directrix is x + y – 9 = 0.

SolutionLet P(x, y) be any point of hyperbola.Let S(2, 2) be the focus. According to the definition of hyperbola

2 2 2SP e PM

22 2 4 x y 9

x 2 y 22

2 2x 4x 4 y 4y 4

2 22 x y 81 2xy 18y 18x

2 2x y 4xy 32x 32y 154 0

This is the required equation of hyperbola.

Class Exercise - 2Find the coordinates of centre, lengths of the axes, eccentricity, length of latus rectum, coordinates of foci, vertices and equation of directrices of the hyperbola

2 2x 2y 2x 8y 1 0

SolutionThe given equation can be written as

2 2x 2x 2 y 4y 1

2 2x 2x 1 2 y 4y 4 6

2 2x 1 2 y 2 6

2 2x 1 y 21

6 3

2 2y 2 x 11 ....(i)

3 6

Solution contd..Shifting the origin at (1, 2) withoutrotating the coordinate axes, i.e. Put x – 1 = X and y – 2 = Y The equation (i) becomes

2 2Y X 1 Conjugate hyperbola

3 6

2 2Here a 6, b 3

Centre: The coordinates of centre with respect to new axes are X = 0 and Y = 0.

The coordinates of centre with respect to oldaxes are x – 1 = 0 and y – 2 = 0.

Solution contd.. x = 1, y = 2 Centre 1, 2

Length of axesLength of transverse axes = 2b

2 3 2 3

Length of conjugate axes = 2a 2 6 2 6

Eccentricity

2 2 2a b e 1

26 3 e 1

2 22 e 1 e 3

e 3

Solution contd..Length of latus rectum

22a 2 6 4 3b 3

Foci: Coordinates of foci with respect to new axes are , i.e. . 0, be X 0, Y 3

Coordinates of foci with respect to old axes are(1, 5) and (1, –1).

Vertices: The coordinates of vertices with respect tonew axes are X = 0 and , i.e. X = 0 and Y b Y 3

Solution contd.. The coordinates of axes with respect toold axes are x – 1 = 0, i.e. x = 1 and

y – 2 3, i. e. 2 3

Vertices 1, 2 3 1, 2 3

Directrices: The equation of directrices with respect

to new axes are , i.e. .bYe Y 1

The equation of directrices with respect to old axes

are , i.e. y = 3 and y = 1. y 2 1

Class Exercise - 3Find the equation of hyperbola whosedirection of axes are parallel tocoordinate axes if(i) vertices are (–8, –1) and (16, –1) and

focus is (17, –1) and(ii) focus is at (5, 12), vertex at (4, 2) and

centre at (3, 2).

Solution(i) Centre of hyperbola is mid-point ofvertices

8 16 1 1Coordinates of centre ,2 2

4, 1

Equation of hyperbola is

2 2

2 2x 4 y 1

1 ...(i)a b

Let x – 4 = X, y + 1 = Y.

Equation of hyperbola in new coordinate axes is

. 2 2

2 2X Y 1a b

Solution contd..As per definition of hyperbolaa = Distance between centre and vertices

2 22a 4 8 1 1

= 144Abscissae of focus in new coordinates system isX = ae, i.e. x – 4 = 12e x 4 12e

17 4 12e x abscissae of focus is 17

13e12 2 2 2b a e 1

169144 1 2514

Solution contd.. Equation of hyperbola is

2 2x 4 y 1

1144 25

2 2i.e. 25x 144y 200x 288y 3344 0

(ii) Coordinates of centre are (3, 2). Equation of hyperbola is

2 2

2 2x 3 y 2

1 ...(i)a b

a = Distance between vertex and centre 2 22i.e. a 4 3 2 2 = 1

Let x – 3 = X, y – 2 = Y. Equation (i) becomes

2 2

2 2X Y 1a b

Solution contd.. Abscissae of focus is X = aei.e. x – 3 = e (As a = 1)

x = e + 3 5 = e + 3 [ Abscissae of focus = 5] e 2

2 2 2b a e 1 = 1 (4 – 1) = 3

2 2x 3 y 2Equation of hyperbola 1

1 3

2 2i.e. 3x y 18x 4y 20 0

Class Exercise - 4Find the equations of the tangents to the hyperbola 4x2 – 9y2 = 36 which are parallel to the line 5x – 3y = 2.

Solution

2 2x yEquation of hyperbola is 1.

9 4

2 2a 9, b 4

Tangent is parallel to the given line 5x – 3y = 2

5Slope of tangent .3

Equation of tangents 2 2 2y mx a m b

5 25i.e. y x 9 43 9

i.e. 5x 3y 3 21

Class Exercise - 5Find the locus of mid-point of portion of tangent intercepted between the axes

for hyperbola 2 2

2 2x y 1.a b

SolutionAny tangent to the hyperbola

is

2 2

2 2x y 1a b

x ysec tan 1 ...(i)a b

Let the tangent (i) intersect the x-axis at A and y-axis at B respectively.Let P(h, k) be the middle point of AB.

Coordinates of A acos , 0 and coordinates of B 0, bcot

P h, k be the middle point of AB

acos 0 2hh cos

2 a

Solution contd..

0 b cot 2kand k cot

2 b

2 2sec tan 1

2 2

2 2a b 14h 4k

2 2

2 2a b Locus of (h, k) is 4.x y

Class Exercise - 6Find the condition that the linelx + my + n = 0 will be normal to the

hyperbola 2 2

2 2x y 1.a b

Solution The equation of the normal at a sec , b tan

2 2

2 2x yto the hyperbola 1 isa b

2 2ax cos by cot a b

2 2i.e. ax sin by a b tan ...(i)

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

Equations (i) and (ii) will represent the same line if

2 2a b tanasin bl m n

Solution contd..

amcosecbl

2 2m a bcot

bn

2 2We know that cosec cot 1

22 2 22 2

2 2 2 2

m a ba m 1b l b n

22 22 2

2 2 2

a ba bl n m

22 22 2

2 2 2

a ba bl m n

Class Exercise - 7

The curve represents

(a) a hyperbola if k < 8(b) an ellipse if k > 8(c) a hyperbola if 8 < k < 12(d) None of these

2 2x y 1,12 k 8 k

SolutionThe given equation

represents hyperbola if(12 – k) (8 – k) < 0

2 2x y 1,12 k 8 k

i.e. 8 < k < 12

Hence, answer is (c).

Class Exercise - 8If the line touches the

hyperbola at the point

, show that

2 2 2y mx a m b

2 2

2 2x y 1a b

asec , b tan

1 bsin .am

Solution Tangent at asec , b tan to the hyperbola

2 2

2 2x y 1isa b

x ysec tan 1 ...(i)a b

sec b bSlope of tangent .

a tan asin

2 2 2Slope of the given line y mx a m b ...(ii) is m.

Solution contd..

Both (i) and (ii) represent same line

bm

asin

bsin

am

1 bsin Proved.am

Class Exercise - 9Let

where be two points on the

hyperbola If (h, k) is the

point of intersection of the normals at P and Q, then k is equal to

(a) (b)

(c) (d)

P asec , b tan and Q asec , b tan ,

2

2 2

2 2x y 1.a b

2 2a ba

2 2a ba

2 2a bb

2 2a bb

Solution

2 2

Equation of normal at P is asec , b tan

ax cos bycot a b ...(i)

2 2

and equation of normal at asec , b tan

is axcos bycot a b ...(ii)

2 2

2 2Equation (ii) reduces to ax sin by tan a b ...(iii)

Solution contd.. h, k be the point of intersection of

normals (i) and (iii),

2 2ahcos bkcot a b ...(iv)

2 2and ahsin bk tan a b ...(v)

From (iv) and (v) eliminating h, we get

2 2cot .sin tan .cos a b sin cos

2 2a bKb

Class Exercise - 10Determine the equations of common tangent to the hyperbola

2 2 2 2

2 2 2 2x y y x1 and 1.a b a b

Solution

2 2

2 2x yEquation of tangent to hyperbola 1a b

2 2 21 1is y m x a m b ....(i)

The equation of other hyperbola can be written as

2 2

2 2x y 1b a

2 2 22 2Equation of tangent is y m x b m a ...(ii)

If equations (i) and (ii) are the same, then 2 2 2 2 2 2

1 2 1 2m m and a m b b m a

Solution contd.. 2 2 2 2 2 2

1 2a m b a b m

2 2 2 2 21 1 2a b m a b m m

21 1m 1 m 1

2 2

Equation of common tangents to both

hyperbola are given by y x a b

Thank you