Question

If the ordinate of any point $P$ on the hyperbola $9x^2-16y^2=144$ is produced to cut the asymptotes in the points $Q$ and $R$. Then the product $PQ.PR$ equals to

• A. $9$
• B. $16$
• C. $25$
• D. $7$

Answer 1

The correct option is A $9$

Hyperbola : $\frac{x^2}{16}-\frac{y^2}{9}=1$

$\Rightarrow a=4,b=3$

let coordinate of $P(4\sec \theta ,3\tan \theta )$

Equations of asymptotes is given by $\frac{x^2}{16}-\frac{y^2}{9}=0$

$\Rightarrow \frac{x}{4}-\frac{y}{3}=0$ and $\frac{x}{4}+\frac{y}{3}=0$

$\Rightarrow y=3/4x$ and $y=-3/4x$

when $x=4\sec \theta y=3\sec \theta$ and $y=-3\sec \theta$

$\Rightarrow Q(4\sec \theta ,3\sec \theta ),R(4\sec \theta -3\sec \theta )$

$\Rightarrow PQ=3\sec \theta -3\tan \theta ,PR=3\tan \theta +3\sec \theta$

$\Rightarrow PQ.PR=(3\sec \theta {)}^2-(3\tan \theta {)}^2$

$=9({\sec }^2\theta -{\tan }^2\theta )=9\Rightarrow \left(A\right)$