Question

A line $lx+my=n$ is normal to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ under the condition $\frac{n^2}{(a^2-b^2{)}^2}(\frac{a^2}{l^2}+\frac{b^2}{m^2})=k,$ then $7k=$

Answer 1

Equation of normal at any point $P(a\cos \theta ,b\sin \theta )$ to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is
$\frac{ax}{\cos \theta }-\frac{by}{\sin \theta }=a^2-b^2\dots \left(i\right)$

Given, line $lx+my=n$ is also normal to ellipse, then there must be a value of $\theta$ for which line $\left(i\right)$ and the given line are identical.
$\therefore \frac{l}{a/\cos \theta }=\frac{m}{-b/\sin \theta }=\frac{n}{a^2-b^2}$
$\Rightarrow \cos \theta =\frac{an}{l(a^2-b^2)},\sin \theta =-\frac{bn}{m(a^2-b^2)}$

We have,
${\sin }^2\theta +{\cos }^2\theta =1\Rightarrow {\left(\frac{an}{l(a^2-b^2)}\right)}^2+{\left(\frac{-bn}{m(a^2-b^2)}\right)}^2=1\Rightarrow \frac{n^2}{(a^2-b^2{)}^2}(\frac{a^2}{l^2}+\frac{b^2}{m^2})=1$

$\therefore k=1\Rightarrow 7k=7$