A hyperbola having the transverse axis of length 2sinθ unit, is confocal with the ellipse 3x2+4y2=12, then its equation is
A
x2cosec2θ−y2sec2θ=1
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B
x2sec2θ−y2cosec2θ=1
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C
x2sin2θ−y2cos2θ=1
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D
x2cos2θ−y2sin2θ=1
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Solution
The correct option is Ax2cosec2θ−y2sec2θ=1 Ellipse is 3x2+4y2=12 ⇒x24+y23=1 b2=a2(1−e2) ⇒e=12 Foucs of ellipse (±ae,0)≡(±1,0) Let equation of confocal hyperbola is x2a21−y2b21=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 x2sin2θ−y2cos2θ=1 ⇒x2cosec2θ−y2sec2θ=1