For ellipse e1=1−a2b2=53
for hyperbola e2=35
Let hyperbola be
a2x2−b2y2=1
∵ it passes through (3,0)⇒a29=1
⇒a2=9
⇒b2=a2(e2−1)
=9(925−1)=16
∴ Hyperbola is
9x2−16y2=1
option (1)
A hyperbola passes through the foci of the ellipse 25x2+16y2=1 and its transverse and conjugate axes coincide with major and minor axes of the ellipse, respectively. If the product of their eccentricities is one, then the equation of the hyperbola is:
Held on 25 Feb 2021 · Verified 6 Jul 2026.
9x2−16y2=1
x2−y2=9
9x2−25y2=1
9x2−4y2=1
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