Question

Let P $(3\sec \theta ,2\tan \theta )$ and $Q(3\sec \varphi ,2\tan \varphi )$ where $\theta +\varphi =\frac{\pi }{2}$, be two distinct points on the hyperbola $\frac{x^2}{9}-\frac{y^2}{4}=1$. Then the ordinate of the point of intersection of the normals at $P$ and $Q$ is:

• A. $\frac{11}{3}$
• B. $-\frac{11}{3}$
• C. $-\frac{13}{2}$
• D. $\frac{13}{2}$

Answer 1

Answer: (C)

$-\frac{13}{2}$

Equation of normal is,
$\frac{a^{2}x}{x_1}+\frac{b^{2}y}{y_1}=a^2+b^2$

At $P$, equation will be

$2\cos \theta x+2\cot \theta y=13$......(1)

At $Q$, equation will be,

$3\sin \theta x+2\tan \theta y=13$......(2)

$\frac{13-2y\tan \theta }{3\sin \theta }=\frac{13-2y\cot \theta }{3\cos \theta }$

$13-2y\tan \theta =13\tan \theta -2y$

$2y-2y\tan \theta =13\tan \theta -13$

$y=-\frac{13}{2}$