AP EAMCET Engineering 2013 Solved Paper

Given, equation of ellipse is

x2

25

+

y2

16

=1
and equation of hyperbola is

x2

4

−

y2

b2

=1
eccentricity of ellipse
b2=a2(1−e2)
⇒16=25(1−e2)
⇒e2=1−

16

25

=

9

25

⇒e=±

3

5

Focii of the ellipse =(±ae,0)=(±3,0) which coincide with focii of the hyperbola.
Let e1 be the eccentricity of the hyperbola.
∴±ae1=±3
⇒e1=

3

2

Now, b2=a2(e12−1)
⇒b2=4(

9

4

−1)=4×

5

4

⇒b2=5