If two points P and Q lie on the hyperbola (x2)/(a2) - (y2)/(b2) =...

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(\dfrac{\cos^{2} \theta}{a^{2}} - \dfrac{\sin^{2} \theta}{b^{2}}\right) =1$<br /><br />Replacing$\thetaby{90}^{o} + $\theta, we get, <br /><br />${r}_{2}^{2}\left(\dfrac{\sin^{2} \theta}{a^{2}} - \dfrac{\cos^{2} \theta}{b^{2}}\right) =1$</p><p>(where,$CQ = {r}_{2}$)<br /><br />$⇒ \dfrac{1}{{r}_{1}^{2}} + \dfrac{1}{{r}_{2}^{2}} = \dfrac{\cos^{2} \theta}{a^{2}} - \dfrac{\sin^{2} \theta}{b^{2}} + \dfrac{\sin^{2} \theta}{a^{2}} - \dfrac{\cos^{2} \theta}{b^{2}}

$⇒ 1{r}_{1}^{2} + 1{r}_{2}^{2} = \cos^{2}$\theta \left(\dfrac{1}{a2} - \dfrac{1}{b2}\right) + \sin^{2} \theta\left(\dfrac{1}{a2} - \dfrac{1}{b2}\right)

$⇒$\dfrac{1}{{r}_{1}^{2}} + \dfrac{1}{{r}_{2}^{2}} = \dfrac{1}{a^{2}} - \dfrac{1}{b^{2}}

$⇒$\dfrac{1}{CP^{2}} + \dfrac{1}{CQ^{2}} = \dfrac{1}{a^{2}} - \dfrac{1}{b^{2}}