IIT JEE Advanced 2006 Paper 1

The given ellipse is

x2

16

+

y2

25

= 1
(b > a) a2 = b2(1−e12)

16

25

= 1−e12 ⇒ e12 = 1 -

16

25

=

9

25

⇒ e1 =

3

5

[focus of ellipse (0, ± be), i.e. (0, ±3)]
e1e2 = 1
Therefore, the eccentricity of hyperbola is
e2 =

5

3

The equation of hyperbola is

x2

A2

−

y2

B2

= - 1
which passes through the focus of the ellipse; therefore,
0 -

9

B2

= - 1 ⇒ B = ± 3
A2 = B2(e22−1)
A2 = 9 (

25

9

−1) = 9(

16

9

) = 16
A = ± 4
Hence, the equation of the given hyperbola is

x2

16

−

y2

9

= - 1