Question

Find the condition that the line $lx+my+n=0$ be a normal to the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1.$

Answer 1

$lx+my+n=0$ ... (1)
Any tangent is $\frac{x\sec \theta }{a}-\frac{y\tan \theta }{b}=1.$
Comparing (1) and (2), we get
$\frac{\sec \theta }{al}=\frac{\tan \theta }{-bm}=\frac{1}{-n}$
$\therefore \sec \theta =-\frac{al}{n},\tan \theta =\frac{bm}{n}$
But ${\sec }^2\theta -{\tan }^2\theta =1$
$\therefore a^2{l}^2-b^2{m}^2=n^2$ is the required condition. Any normal to th hyperbola is
$\frac{ax}{\sec \theta }-\frac{by}{\tan \theta }=a^2+b^2.$ ... (3)
Comparing (1) and (3), we get
$\frac{l\sec \theta }{a}-\frac{m\tan \theta }{b}=\frac{-n}{(a^2+b^2)}$
$\therefore \sec \theta =\frac{a}{l}\left(\frac{-n}{a^2+b^2}\right),\tan \theta =\frac{b}{m}\left(\frac{-n}{a^2+b^2}\right).$
But ${\sec }^2\theta -{\tan }^2\theta =1$
$\therefore {\left(\frac{n}{a^2+b^2}\right)}^2[\frac{a^2}{l^2}-\frac{b^2}{m^2}]=1$
or $\frac{a^2}{l^2}-\frac{b^2}{m^2}=\frac{{(a^2+b^2)}^2}{n^2}.$