IIT JEE Advanced 2007 Paper 1

The given equation of ellipse can be written as

x2

4

+

y2

3

= 1
Now, a = 2 , b = √3 ⇒ e =

1

2

The foci of ellipse is (±1, 0). Since the hyperbola is confocal with ellipse, the foci of Hyperbola is (±1, 0). Now, let e′ be the eccentricity of hyperbola. The transverse axis is
2 sin θ = 2a'
Therefore, the focus is
(e' sin θ , 0) = (1 , 0)
⇒ e' = cosec θ
Now, b′2 = a′2(e′2−1)
= sin2θ(cosec2θ−1)
= 1 - sin2θ
= cos2θ
Therefore, the equation of hyperbola is

x2

sin2θ

−

y2

cos2θ

= 1
That is,
x2cosec2θ–y2sec2θ = 1