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Question

A hyperbola passes through the foci of the ellipse x225+y216=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:

A
x29y24=1
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B
x29y216=1
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C
x2y2=9
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D
x29y225=1
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Solution

The correct option is B x29y216=1
For ellipse e1=11625=35
Foci =(±3,0)
Let equation of hyperbola be x2A2y2B2=1 which passes through (±3,0)
A2=9,A=3,e2=53
Now,
e22=1+B2A2
259=1+B29B2=16

Required equation of hyperbola isx29y216=1

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