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Question

A hyperbola having the transverse axis of length 2sinθ unit, 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
Ellipse is 3x2+4y2=12
x24+y23=1
b2=a2(1e2)
e=12
Foucs of ellipse (±ae,0)(±1,0)
Let equation of confocal hyperbola is
x2a21y2b21=1
and eccentricity =e1
Given, length of transvese axis 2sinθ=2a1
a1=sin θ
Now, focus of hyperbola is same as the ellipse
a1e1=1
e21a21=a21+b21
1=a21+b21
b21=cos2θ
Thus, the equation of hyperbola is
x2sin2θy2cos2θ=1
x2cosec2θy2sec2θ=1

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