Answer:
Option C
Explanation:
Given Ellipse x216+y2b2=1
Now b2=a2(1−e2)
⇒b2=16(1−e2),⇒b216=1−e2
⇒e2=1−b216=16−b216
⇒e=√16−b24
Foci = (±ae,0)=(±√16−b2,0)
Given hyperbola : x214+y281=125
⇒x2(125)2−y2(95)2=1
Now, b2=a2(e2−1)
⇒(95)2=(125)2(e2−1)
⇒(95)2=(e2−1)
⇒e2=1+81144=144+81144
⇒e=1512=54
Foci = (±ae,0)= (±3,0)
Since foci of the given ellipse and hyperbola coincide , therefore
√16−b2=3⇒16−b2=9
∴ b2 =7