Answer:
Option B
Explanation:
Equation of ellipse x216+y225=1
Foci =(0,±3)
⇒ e1=√1−1625=35 given e1e2 =1
(where e2 is eccentricity of hyperbola )
Let equation of hyperbola -x2a2+y2b2=1
its passes through (0, ±3)
∴ b2=9
⇒ e2=√1+a2b2
⇒ e22=1+a2b2
⇒ 1e22=1+a29⇒259=1+a29
⇒ a2=16
Hence, the equation of hyperbola y29−x216=1