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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 e1e2=1, then equation of the hyperbola is


A

x29y216=1

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B

x216y29=1

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C

x29y225=1

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D

None of these

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Solution

The correct option is B

x216y29=1


The eccentricity of x216+y225=1 is
e1=11625=35
e2=53 [e1 e2=1]
Foci of ellipse (0,±3)
Equation of hyperbola is x216y29=1.


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