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Question

If P and Q be two points on the hyperbola x2a2y2b2=1, whose centre is C such that CP is perpnediuclar to CQ,a<b, then the value of 1CP2+1CQ2 is

A
b2a22ab
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B
1a2+1b2
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C
2abb2a2
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D
1a21b2
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Solution

The correct option is D 1a21b2
Given : x2a2y2b2=1


Let CP=r1 be inclined to transverse axis at an angle θ.
P=(r1cosθ,r1sinθ) and P lies on the hyperbola, so
r21(cos2θa2sin2θb2)=1

Replacing θ by 90+θ, we get
Q=(r2sinθ,r2cosθ) and Q also lies on the hyperbola, so
r22(sin2θa2cos2θb2)=1
Now,
1CP2+1CQ2=1r21+1r221CP2+1CQ2=cos2θa2sin2θb2+sin2θa2cos2θb21CP2+1CQ2=1a21b2

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