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Question

If the hyperbola x2b2y2a2=1 passes through the focus of the ellipse x2a2+y2b2=1, then eccentricity of the hyperbola is

A
2
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B
3
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C
3/2
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D
3/2
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Solution

The correct option is D 3
Focusoftheellipsex2a2+y2b2=1willbe(±a2b2,0)Ifthehyperbolapassesthroughthefocusofellipsethen(±a2b2)h2b20=1a2b2=b2[a2=2b2]Eccentricityofthegivenhyperbolacanbegivenase=1+(ab)2=1+2[e=3]

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