KEAM 2013 Math Paper

Given equation of ellipse, ‌

x2

25

+‌

y2

b2

=1.....(i)
and equation of hyperbola, ‌

x2

144

−‌

y2

25

=‌

1

13

⇒‌‌‌

x2

(‌ 144 13 )

−‌

y2

(‌ 25 13 )

=1
Here, A2=‌

144

13

and B2=‌

25

13

∴ Eccentricity of hyperbola is,
∴ Focii of hyperbola is (±Ae2,0)
=(±‌

12

√13

×‌

13

12

,0)=(±√13,0)
Given, focii of the ellipse and hyperbola coincide. i.e., focii of an ellipse (±ae1,0)=(±√13,0)
⇒‌‌ae1=√13
⇒‌‌5e1=√13
⇒‌‌e1=‌

√13

5

Also, e1=√‌

a2−b2

a2

⇒‌‌e12a2=a2−b2
∴‌‌b2=12