If e1 is the eccentricity of the ellipse x216+y225=1 and e2 is the eccentricity of the hyperbola passing through the foci of the ellipse and e1e2=1, then equation of the hyperbola is
x216−y29=−1
The eccentricity of x216+y225=1 is
e1=√1−1625=35
∴e2=53 [∵e1 e2=1]
⇒ Foci of ellipse (0,±3)
⇒ Equation of hyperbola is x216−y29=−1.