Question

If the straight line $lx+my+n=0$ is normal to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ then prove that:
$\frac{a^2}{l^2}+\frac{b^2}{m^2}=\frac{(a^2-b^2{)}^2}{n^2}$

Answer 1

Equation of normal to the given ellipse at $(a\cos \theta ,b\sin \theta )$ is $\frac{ax}{\cos \theta }-\frac{by}{\sin \theta }=a^2-b^2$ ...(1)

If the line $lx+my=n$ is also normal to the ellipse then there must be a value of $\theta$ for which line (1) and line $lx+my=n$ are identical. For that value of $\theta$ we have

$\frac{l}{\frac{a}{\cos \theta }}=\frac{m}{-\frac{b}{\sin \theta }}=\frac{n}{a^2-b^2}$

$⟹\frac{l}{a}\cos \theta =\frac{-m}{b}\sin \theta =\frac{n}{a^2-b^2}$

$⟹\cos \theta =\frac{an}{l(a^2-b^2)}$ and $\sin \theta =\frac{-bn}{m(a^2-b^2)}$

Now, we know

${\cos }^2\theta +{\sin }^2\theta =1$

$⟹{\left(\frac{an}{l(a^2-b^2)}\right)}^2+{\left(\frac{-bn}{m(a^2-b^2)}\right)}^2=1$

$⟹{\left(\frac{a}{l}\right)}^2+{\left(\frac{b}{m}\right)}^2=\frac{(a^2-b^2{)}^2}{n^2}$

Hence proved.