Find the equation of the ellipse whose eccentricity is 4 5 and whose foci coincide with those of the hyperbola 9 x 2 - 16 y 2 + 144 = 0 .

Given the equation of the hyperbola 9 x 2 - 16 y 2 + 144 = 0 . We can write this equation as y 2 9 - x 2 16 = 1 . On comparing the above equation with this equation y 2 b 2 - x 2 a 2 = 1 we get a = 4 & b = 3 As we know the eccentricity of the hyperbola is e = 1 + a 2 b 2 ⇒ e = 1 + 16 9 . ⇒ e = 5 3 So the foci for the equation of the hyperbola with equation y 2 b 2 - x 2 a 2 = 1 , is ( 0 , b e ) & ( 0 , - b e ) So the foci of the hyperbola y 2 9 - x 2 16 = 1 is ( 0 , ± 5 ) . As it is given that the focus of the hyperbola is same as the focus of the ellipse, The eccentricity of the ellipse is also given as 4 5 . As the foci of the ellipse is written as ( 0 , ± b e ) . Thus b = 25 4 As we know the eccentricity of the ellipse a 2 = b 2 ( 1 - e 2 ) . ⇒ a 2 = 225 16 . So the equation of the ellipse is written as 16 x 2 225 + 16 y 2 625 = 1

Find the equation of the ellipse whose eccentricity is 4 5 and whose foci coincide with those of the hyperbola 9 x 2 - 16 y 2 + 144 = 0 .

Given the equation of the hyperbola 9 x 2 - 16 y 2 + 144 = 0 . We can write this equation as y 2 9 - x 2 16 = 1 . On comparing the above equation with this equation y 2 b 2 - x 2 a 2 = 1 we get a = 4 & b = 3 As we know the eccentricity of the hyperbola is e = 1 + a 2 b 2 ⇒ e = 1 + 16 9 . ⇒ e = 5 3 So the foci for the equation of the hyperbola with equation y 2 b 2 - x 2 a 2 = 1 , is ( 0 , b e ) & ( 0 , - b e ) So the foci of the hyperbola y 2 9 - x 2 16 = 1 is ( 0 , ± 5 ) . As it is given that the focus of the hyperbola is same as the focus of the ellipse, The eccentricity of the ellipse is also given as 4 5 . As the foci of the ellipse is written as ( 0 , ± b e ) . Thus b = 25 4 As we know the eccentricity of the ellipse a 2 = b 2 ( 1 - e 2 ) . ⇒ a 2 = 225 16 . So the equation of the ellipse is written as 16 x 2 225 + 16 y 2 625 = 1

Prove that the asymptotes to the hyperbola x 2 a 2 - y 2 b 2 = 1 are the tangents from its centre.

We know that the equation of pair of tangents drawn from the point P x 1 , y 1 to the hyperbola x 2 a 2 - y 2 b 2 = 1 is x 2 a 2 - y 2 b 2 x 1 2 a 2 - y 1 2 b 2 = x 1 x a 2 - y 1 y b 2 2 . And we also know that the equation of asymptotes of the hyperbola x 2 a 2 - y 2 b 2 = 1 are y = ± b a x . Now, the equation of pair of tangents drawn from the centre 0 , 0 to the hyperbola x 2 a 2 - y 2 b 2 = 1 is x 2 a 2 - y 2 b 2 - 1 0 2 a 2 - 0 2 b 2 - 1 = 0 . x a 2 - 0 . y b 2 - 1 2 . ⇒ - x 2 a 2 + y 2 b 2 + 1 = 1 ⇒ y 2 = b 2 a 2 x 2 ⇒ y = ± b a x The above equations are the asymptotes of the hyperbola x 2 a 2 - y 2 b 2 = 1 .

Find the locus of the points of intersection of two tangents to a hyperbola x 2 25 - y 2 16 = 1 . If sum of their slopes is constant a

Given the equation of the hyperbola x 2 25 - y 2 16 = 1 . Now comparing this equation with the standard equation of the hyperbola we get a = 5 & b = 4 . Let the slope of the equation of two tangents to the hyperbola be m 1 & m 2 . As we know the equation of the tangent to the hyperbola is written as y = m x ± a 2 m 2 - b 2 . ⇒ y = m x ± 25 m 2 - 16 . Let the coordinates of the point of the intersection of two tangents to a hyperbola be ( h , k ) . ⇒ k - m h = ± 25 m 2 - 16 . Squaring both side we get. ⇒ k 2 + m 2 h 2 - 2 m k h = 25 m 2 - 16 . ⇒ m 2 ( h 2 - 25 ) - 2 m h k + k 2 + 16 = 0 . As the equation is in the form of the quadratic equation a x 2 + b x + c = 0 , so the sum of the roots of this quadratic equation is - b a . So we can write m 1 + m 2 = - - 2 h k ( h 2 - 25 ) ⇒ m 1 + m 2 = 2 h k h 2 - 25 It is given that m 1 + m 2 = a . ⇒ 2 h k h 2 - 25 = a ⇒ a ( h 2 - 25 ) = 2 h k . So the locus of the point of intersection of the tangents is a ( x 2 - 25 ) = 2 x y .

Prove that the locus of the intersection of tangents to the hyperbola x 2 a 2 - y 2 b 2 = 1 , which meet at a constant angle β , is the curve x 2 + y 2 + b 2 - a 2 2 = 4 cot 2 ⁡ β a 2 y 2 - b 2 x 2 + a 2 b 2 .

Given the equation to the hyperbola, x 2 a 2 - y 2 b 2 = 1 . . . 1 The equation of any tangent of ( 1 ) is y = m x ± a 2 m 2 - b 2 . . . 2 If this tangent passes through a point ( h , k ) , then we get k = m h ± a 2 m 2 - b 2 ⇒ k - m h = ± a 2 m 2 - b 2 Squaring both sides we get, k - m h 2 = ± a 2 m 2 - b 2 2 ⇒ k - m h 2 = a 2 m 2 - b 2 On simplifying we get a quadratic equation in m , m 2 h 2 - a 2 - 2 m h k + k 2 + b 2 = 0 . . . ( 3 ) The equation has ( 3 ) has two roots, let m 1 and m 2 be the roots, then m 1 + m 2 = 2 h k h 2 - a 2 . . . 4 and m 1 m 2 = k 2 + b 2 h 2 - a 2 . . . 5 Given angle between two tangents is constant β . So applying the formula of angle between two lines we get, tan β = m 1 - m 2 1 + m 1 m 2 ⇒ tan β = m 1 + m 2 2 - 4 m 1 m 2 1 + m 1 m 2 (Since m 1 - m 2 = m 1 + m 2 2 - 4 m 1 m 2 ) ⇒ cot β = 1 + m 1 m 2 m 1 + m 2 2 - 4 m 1 m 2 . . . 6 Squaring both sides and using equations ( 4 ) and ( 5 ) , we get cot 2 β m 1 + m 2 2 - 4 m 1 m 2 = 1 + m 1 m 2 2 ⇒ cot 2 β 2 h k h 2 - a 2 2 - 4 k 2 + b 2 h 2 - a 2 = 1 + k 2 + b 2 h 2 - a 2 2 ⇒ cot 2 β 4 h 2 k 2 - 4 h 2 - a 2 k 2 + b 2 = h 2 - a 2 + k 2 + b 2 2 ⇒ h 2 + k 2 + b 2 - a 2 2 = 4 cot 2 β a 2 k 2 - b 2 h 2 + a 2 b 2 Generalizing, the required locus is x 2 + y 2 + b 2 - a 2 2 = 4 cot 2 ⁡ β a 2 y 2 - b 2 x 2 + a 2 b 2 2 . Hence proved.

EASY

JEE Main/Advanced

IMPORTANT

Earn 100

Find the equation of the ellipse whose eccentricity is $\frac{4}{5}$ and whose foci coincide with those of the hyperbola $9 x^{2} - 16 y^{2} + 144 = 0$ .

Important Questions on Hyperbola

MEDIUM

JEE Main/Advanced

IMPORTANT

Skills in Mathematics Coordinate Geometry for JEE Main & Advanced>Chapter 7 - Hyperbola>EXERCISE ON LEVEL-I>Q 6

The hyperbola

\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1

passes through the point of intersection of the lines,

7 x + 13 y - 87 = 0

and

5 x - 8 y + 7 = 0,

the latus rectum is

\frac{32 \sqrt{2}}{5} .

Find

a & b

MEDIUM

JEE Main/Advanced

IMPORTANT

Skills in Mathematics Coordinate Geometry for JEE Main & Advanced>Chapter 7 - Hyperbola>EXERCISE ON LEVEL-I>Q 7

PN

is the ordinate of any point

P

on the hyperbola

\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1 .

Q

divides

AP

in the ratio

a^{2} : b^{2},

show that

NQ

is perpendicular to

A^{'} P

where

{AA}^{'}

is the transverse axis of the hyperbola.

HARD

JEE Main/Advanced

IMPORTANT

Skills in Mathematics Coordinate Geometry for JEE Main & Advanced>Chapter 7 - Hyperbola>EXERCISE ON LEVEL-I>Q 9

A normal to the hyperbola

x^{2} - 4 y^{2} = 4

meets the

x

and

y

axes at

A

and

B

respectively. Find the locus of the point of intersection of the straight lines drawn through

A

and

B

perpendicular to the

x

and

y

axes respectively.

EASY

JEE Main/Advanced

IMPORTANT

Skills in Mathematics Coordinate Geometry for JEE Main & Advanced>Chapter 7 - Hyperbola>EXERCISE ON LEVEL-I>Q 10

A point

P

moves such that the tangents

P T_{1}

and

P T_{2}

from it to the hyperbola

4 x^{2} - 9 y^{2} = 36

are mutually perpendicular. Then the equation of the locus of

P

HARD

JEE Main/Advanced

IMPORTANT

Skills in Mathematics Coordinate Geometry for JEE Main & Advanced>Chapter 7 - Hyperbola>EXERCISE ON LEVEL-I>Q 11

Prove that the asymptotes to the hyperbola

\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1

are the tangents from its centre.

EASY

JEE Main/Advanced

IMPORTANT

Skills in Mathematics Coordinate Geometry for JEE Main & Advanced>Chapter 7 - Hyperbola>EXERCISE ON LEVEL-I>Q 12

Find the locus of the points of intersection of two tangents to a hyperbola

\frac{x^{2}}{25} - \frac{y^{2}}{16} = 1 .

If sum of their slopes is constant

a

HARD

JEE Main/Advanced

IMPORTANT

Skills in Mathematics Coordinate Geometry for JEE Main & Advanced>Chapter 7 - Hyperbola>EXERCISE ON LEVEL-I>Q 13

Prove that the locus of the intersection of tangents to the hyperbola

\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1

, which meet at a constant angle

β

, is the curve

{(x^{2} + y^{2} + b^{2} - a^{2})}^{2} = 4 \cot^{2} β (a^{2} y^{2} - b^{2} x^{2} + a^{2} b^{2})

EASY

JEE Main/Advanced

IMPORTANT

Skills in Mathematics Coordinate Geometry for JEE Main & Advanced>Chapter 7 - Hyperbola>EXERCISE ON LEVEL-I>Q 16

Show that the locus represented by

x = \frac{1}{2} a (t + \frac{1}{t}), y = \frac{1}{2} a (t - \frac{1}{t})

is a rectangular hyperbola. Show also that the equation to the normal at the point '

t

' is

\frac{x}{t^{2} + 1} + \frac{y}{t^{2} - 1} = \frac{a}{t}

Find the equation of the ellipse whose eccentricity is 45 and whose foci coincide with those of the hyperbola 9x2-16y2+144=0.

Important Questions on Hyperbola

Learn and Prepare for any exam you want !

Find the equation of the ellipse whose eccentricity is $\frac{4}{5}$ and whose foci coincide with those of the hyperbola $9 x^{2} - 16 y^{2} + 144 = 0$ .