Question

If C is the centre of the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ and the tangent 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 is equal to:

• A. $a^2+b^2$
• B. $|a^2-b^2|$
• C. $b^2/a^2$
• D. $a^2/b^2$

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 $bx-ay=a$ $\Rightarrow \frac{x}{a}=\frac{y}{b}$ 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=$$(\frac{a}{\sec \theta +\tan \theta }.\frac{-b}{\sec \theta +\tan \theta })$
$\therefore$ $CQ-CR$$=\frac{\sqrt{a^2+b^2}}{(\sec \theta -\tan \theta )}.\frac{\sqrt{a^2+b^2}}{(\sec \theta +\tan \theta )}$
$=a^2+b^2$