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

Let $CP=r_1$ be inclined to transverse axis at an angle $\theta$ such that $P$$\left(r_1\cos \theta , r_1\sin \theta \right)$ lies on the hyperbola. Therefore,

$r_1^2 \left(\frac{{\cos }^2\theta }{a^2}-\frac{{\sin }^2\theta }{b^2}\right)=1$

Replacing $\theta by {90}^o+\theta ,$ we get, 

$r_2^2 \left(\frac{{\sin }^2\theta }{a^2}-\frac{{\cos }^2\theta }{b^2}\right)=1$

(where, $CQ=r_2$)

$\Rightarrow \frac{1}{r_1^2}+\frac{1}{r_2^2}=\frac{{\cos }^2\theta }{a^{2 }}-\frac{{\sin }^2\theta }{b^2}+\frac{{\sin }^2\theta }{a^2}-\frac{{\cos }^2\theta }{b^2}$

$\Rightarrow \frac{1}{r_1^2}+\frac{1}{r_2^2}={\cos }^2\theta \left(\frac{1}{a^2}-\frac{1}{b^2}\right)+{\sin }^2\theta \left(\frac{1}{a^2}-\frac{1}{b^2}\right)$

$\Rightarrow \frac{1}{r_1^2}+\frac{1}{r_2^2}=\frac{1}{a^2}-\frac{1}{b^2}$

$\Rightarrow \frac{1}{C{P}^2}+\frac{1}{C{Q}^2}=\frac{1}{a^2}-\frac{1}{b^2}$