COMEDK Exam 2013 Solved Paper

The equation of the ellipse is

x2

16

+

y2

b2

=1 and
equation of the hyperbola is

x2

144

−

y2

81

=

1

25

or

x2

( 12 5 )2

−

y

( 9 5 )2

=1
∴ Eccemtricity of ellipe, e=√1−

b2

16

and eccentricity of hyperbola,
e′=√1+

( 9 5 )2

( 12 5 )2

=√1+

81

144

=

15

12

Since, foci of ellipse and hyperbola coincide.
∴4e=

12

5

e′
⇒e=

3

5

e′
⇒√1−

b2

16

=

3

5

×

15

12

⇒1−

b2

16

=(

3

4

)2⇒b2=7