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Question

A hyperbola having the transverse axis of length 2sinθ, is confocal with the ellipse 3x2+4y2=12. Then its equation is :

A
x2cosec2θy2sec2θ=1
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B
x2sec2θy2cosec2θ=1
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C
x2sec2θy2cos2θ=1
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D
x2cos2θy2sin2θ=1
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Solution

The correct option is A x2cosec2θy2sec2θ=1
The given ellipse is
x24+y23=1
a=2,b=3
3=4(1e2)
e=12
So that , ae=1
Hence , the eccentricity e1 of the hyperbola is given by
1=e1sinθ
e1=cosecθ
b2=sin2θ(cosec2θ1)
b2=cos2θ
Hence , the Hyperbola is
x2sin2θy2cos2θ=1
or
x2cosec2θy2sec2θ=1

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