$If two points P and Q on the hyperbola \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1 whose centre C be such that CP is perpendicular to CQ, a < b, then the value of \frac{1}{{CP}^{2}} + \frac{1}{{CQ}^{2}} is$

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$\frac{b^{2} - a^{2}}{2 ab}$

$\frac{1}{a^{2}} + \frac{1}{b^{2}}$

$\frac{2 ab}{b^{2} - a^{2}}$

$\frac{1}{a^{2}} - \frac{1}{b^{2}}$

answer is D.

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Detailed Solution

$Let CP = r_{1}, ∠ XC P = θ ∴ P = (r_{1} \cos θ, r_{1} \sin θ) lie on hyperbola \Rightarrow \frac{1}{r_{1}^{2}} = \frac{\cos^{2} θ}{a^{2}} - \frac{\sin^{2} θ}{b^{2}} . . . . . (1) Let CQ = r_{2}, ∠ XC Q = \frac{π}{2} + θ ∴ Q = (- r_{2} \sin θ, r_{2} \cos θ) lie on hyperbola \begin{array}{l} \Rightarrow \frac{1}{r_{2}^{2}} = (\frac{\sin^{2} θ}{a^{2}} - \frac{\cos^{2} θ}{b^{2}}) \dots \dots \dots . . (2) \\ (1) + (2) \\ \Rightarrow \frac{1}{r_{1}^{2}} + \frac{1}{r_{2}^{2}} = \frac{1}{a^{2}} - \frac{1}{b^{2}} \\ \Rightarrow \frac{1}{{CP}^{2}} + \frac{1}{{CQ}^{2}} = \frac{1}{a^{2}} - \frac{1}{b^{2}} \end{array}$