Question

# 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 e1.e2=1, then the equation of the hyperbola is

A
16x281y29=1
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B
16x281y29=1
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C
x2916y281=1
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D
x2916y281=1
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Solution

## The correct option is A 16x281−y29=−1 x216+y225=1 e1=√25−1625=35 e1.e2=1 ⇒e2=53 Equation of hyperbola : x2a2−y2b2=1 (Assume) e22=1+b2a2⇒b2=169a2 ∴x2a2−y2169a2=1 Hyperbola passes through the foci of ellipse (0,±3) ⇒−9169a2=1⇒a2=−8116 So, our assumption is wrong. Therefore, equation of hyperbola is x2a2−y2b2=−1 ⇒x2(8116)−y2169.8116=−1 ⇒16x281−y29=−1

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