Question

Let $P(a\sec \theta ,b\tan \theta )$ and $Q(a\sec \varphi ,b\tan \varphi )$ where $\theta +\varphi =\frac{\pi }{2}$ be two points on the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$. if $(h,k)$ is the point of intersection of the normals at $P$ and $Q$ then $k$ is

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

Answer 1

The correct option is D $-\frac{a^2+b^2}{b}$

Equations of the normal at P is $ax+bycosec\theta =(a^2+b^2)\sec \theta$ (i)
and the equation of the normal at $Q(a\sec \varphi ,b\sec \varphi )$ is
$ax+bycosec\varphi =(a^2+b^2)\sec \varphi$ (ii)
Subtracting (ii) from (i) we get
$y=\frac{a^2+b^2}{b}.\frac{\sec \theta -\sec \varphi }{cosec\theta -cosec\varphi }$
so $k=y=\frac{a^2+b^2}{b}.\frac{\sec \theta -\sec (\pi /2-\theta )}{cosec\theta -cosec(\pi /2-\theta )}$ $[\because \theta +\varphi =\pi /2]$
$=\frac{a^2+b^2}{b}.\frac{\sec \theta -cosec\theta }{cosec\theta -\sec \theta }=-\left[\frac{a^2+b^2}{b}\right]$
Hence, option 'D' is correct