Question

Let P $\left(3\, sec \,\theta, \,2 \,tan \,\theta\right)$ and Q $\left(3\, sec\, \phi, \,2 \,tan \,\phi\right)$ where $\theta + \phi = \frac{\pi}{2},$ 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

The Correct Option is D

$p \left(3 \,sec\, \theta,\,2 \,tan \,\theta\right) \quad Q = \left(3 \,sec\,\phi\, ,\, 2 \,tan\phi\right)$
$\theta +\phi = \frac{\pi}{2}\quad Q = \left(3 \,cosec \, \theta , \,2 \, cot\theta\right)$
Equation of normal at $p =$
$= 3x \, cos \,\theta + 2y \, cot \,\theta=13$
$= 3x \, sin \,\theta \, cos \,\theta + 2y \,cos \,\theta = 13 \,sin\theta\quad ...\left(1\right)$
equation of normal at Q $\Rightarrow$
$= 3x\, sin \, \theta + 2y \, tan \, \theta = 13$
$= 3x\, sin\, \theta \, cos \, \theta + 2y \, sin \, \theta = 13\, cos\theta \quad ...\left(2\right)$
$\left(1\right)-\left(2\right)\, \Rightarrow$
$2y \, \left(cos\, \theta - sin \, \theta \right) = 13 \left(sin\, \theta - cos \,\theta \right)$
$2y = -13 \Rightarrow \frac{-13}{2}$