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 normals at $P$ & $Q$, then find $k$.

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

Answer 1

The correct option is A $-\left(\frac{a^2+b^2}{b^2}\right)$

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 'A' is correct.