Question

Prove that the straight line $lx+my=n$ is a normal to the ellipse, if $\frac{a^2}{l^2}+\frac{b^2}{m^2}=\frac{(a^2-b^2{)}^2}{n^2}$.

Answer 1

We have,

Equation of straight line is

$lx+my=n......\left(1\right)$

Normal to the ellipse

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1(a>b)$

We know that equation of line

$y=mx+c$

Equation of line normal to the ellipse be

$y=mx\pm \frac{m(a^2-b^2)}{\sqrt{a^2+b^2{m}^2}}$

On comparing that,

$c=\pm \frac{m(a^2-b^2)}{\sqrt{a^2+b^2{m}^2}}......\left(2\right)$

From (1),

$lx+my+n=0$

$\Rightarrow y=-\frac{l}{m}x-\frac{n}{m}$

Now,

$y=mx+c$

On comparing that,

$c=\frac{-n}{m}......\left(3\right)$

$slope=\frac{-l}{m}$

From (2) and (3) to, and we get,

$\pm \frac{m(a^2-b^2)}{\sqrt{a^2+b^2{m}^2}}=\frac{-n}{m}$

On squaring both side and we get,

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

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

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

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

Hence proved.