Question

$C$ is the center of the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$. The tangents at any point $P$ on this hyperbola meets the straight lines $bx-ay=0$ and $bx+ay=0$ in the points $Q$ and $R$ respectively. Then $CQ.CR=$

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

Answer 1

The correct option is A $a^2+b^2$

The coordinates of the point $P$ are $(a\sec \theta ,b\tan \theta )$

Tangent at $P$ is $\frac{x\sec \theta }{a}-\frac{y\tan \theta }{b}=1$

It meets $bx-ay=0\Rightarrow \frac{x}{a}=\frac{y}{b}$ in $Q$

$\therefore Q$ is $(\frac{a}{\sec \theta -\tan \theta },\frac{-b}{\sec \theta -\tan \theta })$

It meets $bx+ay=0\Rightarrow \frac{x}{a}=-\frac{y}{b}$ in $R$.

$\therefore R$ is $(\frac{a}{\sec \theta +\tan \theta },\frac{-b}{\sec \theta +\tan \theta })$

$\therefore CQ.CR=\frac{\sqrt{a^2+b^2}}{(\sec \theta -\tan \theta )}.\frac{\sqrt{a^2+b^2}}{(\sec \theta -\tan \theta )}=a^2+b^2(\because {\sec }^2\theta -{\tan }^2\theta =1)$