If two points $P$ and $Q$ on the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ whose center is $C$ be such that $CP$ is perpendicular to $CQ$ and $a<b$,then prove that $\frac{1}{CP^2}+\frac{1}{CQ^2}=\frac{1}{a^2}-\frac{1}{b^2}$.
I let $P(a\sec\alpha,b\tan\alpha)$ and $Q(a\sec\beta,b\tan\beta)$,Center $C$ is $(0,0)$.
$CP^2=a^2\sec^2\alpha+b^2\tan^2\alpha$ and $CQ^2=a^2\sec^2\beta+b^2\tan^2\beta$
We need to prove that $\frac{1}{a^2\sec^2\alpha+b^2\tan^2\alpha}+\frac{1}{a^2\sec^2\beta+b^2\tan^2\beta}=\frac{1}{a^2}-\frac{1}{b^2}……….(1)$
We are given that slope of $CQ\times$ slope of $CP=-1$
$\frac{b\tan\beta}{a\sec\beta}\times\frac{b\tan\alpha}{a\sec\alpha}=-1……….(2)$
I am not able to prove equation $(1)$ from the given equation $(2.)$
Best Answer
Using $$\begin{align}(2)&\Rightarrow \sin\alpha=-\frac{a^2}{b^2\sin\beta}\\\\&\Rightarrow 1-\cos^2\alpha=\frac{a^4}{b^4\sin^2\beta}\\\\&\Rightarrow \cos^2\alpha=1-\frac{a^4}{b^4\sin^2\beta}=\frac{b^4\sin^2\beta-a^4}{b^4\sin^2\beta}\end{align}$$ and eliminating $\alpha$ to obtain $$\begin{align}CP^2&=a^2\sec^2\alpha+b^2\tan^2\alpha\\\\&=a^2\cdot\frac{b^4\sin^2\beta}{b^4\sin^2\beta-a^4}+b^2\cdot\frac{a^4}{b^4\sin^2\beta-a^4}\\\\&=\frac{a^2b^2(b^2\sin^2\beta+a^2)}{b^4\sin^2\beta-a^4}\end{align}$$ we get $$\begin{align}\frac{1}{CP^2}+\frac{1}{CQ^2}&=\frac{b^4\sin^2\beta-a^4}{a^2b^2(b^2\sin^2\beta+a^2)}+\frac{1}{a^2\sec^2\beta+b^2\tan^2\beta}\\\\&=\frac{b^4\sin^2\beta-a^4}{a^2b^2(b^2\sin^2\beta+a^2)}+\frac{\cos^2\beta}{a^2+b^2\sin^2\beta}\\\\&=\frac{b^4\sin^2\beta-a^4}{a^2b^2(b^2\sin^2\beta+a^2)}+\frac{a^2b^2\cos^2\beta}{a^2b^2(a^2+b^2\sin^2\beta)}\\\\&=\frac{b^4\sin^2\beta-a^4+a^2b^2(1-\sin^2\beta)}{a^2b^2(b^2\sin^2\beta+a^2)}\\\\&=\frac{(b^4-a^2b^2)\sin^2\beta-a^4+a^2b^2}{a^2b^2(b^2\sin^2\beta+a^2)}\\\\&=\frac{b^2(b^2-a^2)\sin^2\beta+a^2(b^2-a^2)}{a^2b^2(b^2\sin^2\beta+a^2)}\\\\&=\frac{(b^2-a^2)(b^2\sin^2\beta+a^2)}{a^2b^2(b^2\sin^2\beta+a^2)}\\\\&=\frac{b^2-a^2}{a^2b^2}\\\\&=\frac{1}{a^2}-\frac{1}{b^2}\end{align}$$