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Question

# P is point on ellipse x2a2+y2b2=1. Its polar line w.r.t. hyperbola x2a2−y2b2=1 meets the asymptotes of the hyperbola in Q and R. If QR=c, then the value of eccentric angle θ of P is given by:

A
c2cos22θ+2(a2b2)cos22θ=2(a2+b2)
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B
c2cos2θ+2(a2b2)cosθ=2(a2+b2)
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C
c2cos2θ+2(a2b2)cos2θ=2(a2+b2)
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D
c2cos22θ+2(a2b2)cos2θ=2(a2+b2)
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Solution

## The correct option is C c2cos22θ+2(a2−b2)cos2θ=2(a2+b2)Let the point P be (acosθ,bsinθ)So, the polar line w.r.t hyperbola will be:xacosθa2−ybsinθb2=1 Or, xcosθa−ysinθb=1 The asymptotes of the hyperbola is: y=±bxa The point of intersection Q will be: xcosθa−bxsinθab=1 Or, x=acosθ−sinθ and y=bcosθ−sinθAnd, the point of intersection R will be: xcosθa+bxsinθab=1Or, x=acosθ+sinθ and y=−bcosθ+sinθ Given, QR=c Or a2(cosθ+sinθ−cosθ+sinθ)2((cosθ)2−(sinθ)2)+b2(cosθ+sinθ+cosθ−sinθ)2((cosθ)2−(sinθ)2)=c2Or 4a2(sinθ)2+4b2(cosθ)2=c2(cos2θ)2 Or 4b2+4(a2−b2)(sinθ)2=c2(cos2θ)2 Or −4b2−4(a2−b2)(sinθ)2=−c2(cos2θ)2 Adding and subtracting 2(a2−b2), we get −4b2−2(a2−b2)+2(a2−b2)(1−2(sinθ)2)=−c2(cos2θ)22(−a2−b2)+2(a2−b2)(cos2θ)=−c2(cos2θ)2 c2(cos2θ)2+2(a2−b2)(cos2θ)=2(a2+b2)

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