Q.

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 eccentricites is one,then the equation of the hyperbola is

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a

x29-y216=1

b

x2-y2=9

c

x29-y225=1

d

x29-y24=1

answer is C.

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Detailed Solution

Given ellipse x225+y216=1(a>b) e=25-1625=35 foci=(±ae, 0)=(3, 0) let equation of hyperbola be x2a2-y2b2=1 It passes through (3, 0) then 9a2=1  a2=9 Given e1×e2=1  35×e2=1                                e2=53                                e22=259                                1+b2a2=259                                b29=169                                b2=16 Hyperbola is x29-y216=1

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