If P and Q be two points on the hyperbola x2a2−y2b2=1, whose centre is C such that CP is perpnediuclar to CQ,a<b, then the value of 1CP2+1CQ2 is
A
b2−a22ab
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B
1a2+1b2
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C
2abb2−a2
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D
1a2−1b2
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Solution
The correct option is D1a2−1b2 Given : x2a2−y2b2=1
Let CP=r1 be inclined to transverse axis at an angle θ. P=(r1cosθ,r1sinθ) and P lies on the hyperbola, so r21(cos2θa2−sin2θb2)=1
Replacing θ by 90∘+θ, we get Q=(−r2sinθ,r2cosθ) and Q also lies on the hyperbola, so r22(sin2θa2−cos2θb2)=1
Now, 1CP2+1CQ2=1r21+1r22⇒1CP2+1CQ2=cos2θa2−sin2θb2+sin2θa2−cos2θb2∴1CP2+1CQ2=1a2−1b2