Question

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

• A. $\frac{b^2-a^2}{2ab}$
• B. $\frac{1}{a^2}+\frac{1}{b^2}$
• C. $\frac{2ab}{b^2-a^2}$
• D. $\frac{1}{a^2}-\frac{1}{b^2}$

Answer 1

The correct option is D $\frac{1}{a^2}-\frac{1}{b^2}$

Given : $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$


Let $CP=r_1$ be inclined to transverse axis at an angle $\theta$.
$P=(r_1\cos \theta ,r_1\sin \theta )$ and $P$ lies on the hyperbola, so
$r_1^2(\frac{{\cos }^2\theta }{a^2}-\frac{{\sin }^2\theta }{b^2})=1$

Replacing $\theta$ by ${90}^{\circ }+\theta$, we get
$Q=(-r_2\sin \theta ,r_2\cos \theta )$ and $Q$ also lies on the hyperbola, so
$r_2^2(\frac{{\sin }^2\theta }{a^2}-\frac{{\cos }^2\theta }{b^2})=1$
Now,
$\frac{1}{C{P}^2}+\frac{1}{C{Q}^2}=\frac{1}{r_1^2}+\frac{1}{r_2^2}\Rightarrow \frac{1}{C{P}^2}+\frac{1}{C{Q}^2}=\frac{{\cos }^2\theta }{a^2}-\frac{{\sin }^2\theta }{b^2}+\frac{{\sin }^2\theta }{a^2}-\frac{{\cos }^2\theta }{b^2}\therefore \frac{1}{C{P}^2}+\frac{1}{C{Q}^2}=\frac{1}{a^2}-\frac{1}{b^2}$