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

Let $\mathrm{CP}={\mathrm{r}}_1$ be inclined to transverse axis at an angle $\theta$ so that

P is $\left({\mathrm{r}}_1\cos \mathrm{\theta },{\mathrm{r}}_1\sin \mathrm{\theta }\right)$ and P lies on the hyperbola

$\Rightarrow {\mathrm{r}}_1^2\left(\frac{{\cos }^2\mathrm{\theta }}{{\mathrm{a}}^2}-\frac{{\sin }^2\mathrm{\theta }}{{\mathrm{b}}^2}\right)=1$

Replacing $\mathrm{\theta }$ by ${90}^0+\mathrm{\theta }$, then 

$\begin{array}{l}{\mathrm{r}}_2^2\left(\frac{{\sin }^2\mathrm{\theta }}{{\mathrm{a}}^2}-\frac{{\cos }^2\mathrm{\theta }}{{\mathrm{b}}^2}\right)=1\\ \Rightarrow \frac{1}{{\mathrm{r}}_1^2}+\frac{1}{{\mathrm{r}}_2^2}=\frac{{\cos }^2\mathrm{\theta }}{{\mathrm{a}}^2}-\frac{{\sin }^2\mathrm{\theta }}{{\mathrm{b}}^2}+\frac{{\sin }^2\mathrm{\theta }}{{\mathrm{a}}^2}-\frac{{\cos }^2\mathrm{\theta }}{{\mathrm{b}}^2}\\ \Rightarrow \frac{1}{{\mathrm{r}}_1^2}+\frac{1}{{\mathrm{r}}_2^2}={\cos }^2\mathrm{\theta }\left(\frac{1}{{\mathrm{a}}^2}-\frac{1}{{\mathrm{b}}^2}\right)+{\sin }^2\mathrm{\theta }\left(\frac{1}{{\mathrm{a}}^2}-\frac{1}{{\mathrm{b}}^2}\right)\\ \Rightarrow \frac{1}{{\mathrm{r}}_1^2}+\frac{1}{{\mathrm{r}}_2^2}=\frac{1}{{\mathrm{a}}^2}-\frac{1}{{\mathrm{b}}^2}\Rightarrow \frac{1}{{\mathrm{CP}}^2}+\frac{1}{{\mathrm{CQ}}^2}=\frac{1}{{\mathrm{a}}^2}-\frac{1}{{\mathrm{b}}^2}\end{array}$