VITEEE 2016 Paper

Given ellipse

x2

16

+

y2

b2

= 1
Now b2 = a2(1−e2)
⇒ b2 = 16 (1−e2) , ⇒

b2

16

= 1 - e2
⇒ e2 = 1 -

b2

16

=

16−b2

16

⇒ e =

√16−b2

4

Foci = (≠ ae , 0) = (±√16−b2,0)
Given hyperbola :

x2

144

−

y2

81

=

1

25

⇒

x2

( 12 5 )2

−

y2

( 9 5 )2

= 1
Now, b2 = a2(e2−1)
⇒ (

9

5

)2 = (

12

5

)2 (e2 - 1)
⇒ (

9

12

)2 = e2 - 1
⇒ $$e^2 = 1 +

81

144

=

144+81

144

⇒ e =

15

12

=

5

4

Foci = (≠ ae , 0) = (≠ 3 , 0)
Since foci of the given ellipse and hyperbola coincide, therefore √16−b2 = 3 ⇒ 16 - b2 = 9
∴ b2 = 7