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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
x2sin2θy2cos2θ=1
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D
x2cos2θy2sin2θ=1
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Solution

The correct option is A x2cosec2θy2sec2θ=1
Given ellipse may be written as x24+y23=1
a2=4,b2=3
e=134=12
Focus of the ellipse =(±ae,0)=(±1,0)
Given required hyperbola is confocal to the ellipse
Let a,b,e are transverse axis, conjugate axis an eccentricity of the hyperbola
ae=1sinθ.e=1e=1sinθ
Using b2=a2(e21)b2=1sin2θ=cos2θ
Therefore required hyperbola is x2a2y2b2=1
x2sin2θy2cos2θ=1
x2cosec2θy2sec2θ=1

Hence, option 'A' is correct.

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