Question

If line $lx+my+n=0$ cuts the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ at points whose eccentric angles differ by $\frac{\pi }{2}$, then the value of $\frac{a^2{l}^2+b^2{m}^2}{n^2}$ is

Answer 1

Let eccentric angles are $\theta$ and $\varphi ,$ then
$\theta -\varphi =\frac{\pi }{2}$ (given)
$\Rightarrow \theta =\frac{\pi }{2}+\varphi$
The line joining the points $\theta$ and $\varphi$ is
$\frac{x}{a}\cos \left(\frac{\theta +\varphi }{2}\right)+\frac{y}{b}\sin \left(\frac{\theta +\varphi }{2}\right)=\cos \left(\frac{\theta -\varphi }{2}\right)$
$\because (\theta =\frac{\pi }{2}+\varphi )$
$\Rightarrow \frac{x}{a}\cos (\frac{\pi }{4}+\varphi )+\frac{y}{b}\sin (\frac{\pi }{4}+\varphi )=\cos \frac{\pi }{4}$
$\Rightarrow \frac{x}{a}\cos (\frac{\pi }{4}+\varphi )+\frac{y}{b}\sin (\frac{\pi }{4}+\varphi )=\frac{1}{\sqrt{2}}...\left(1\right)$

The given line is $lx+my+n=0$
$\Rightarrow lx+my=-n...\left(2\right)$

Now, equation $\left(1\right)$ and $\left(2\right)$ represent the same line, so comparing them, we get
$\frac{\cos (\frac{\pi }{4}+\varphi )}{la}=\frac{\sin (\frac{\pi }{4}+\varphi )}{mb}=-\frac{1}{n\sqrt{2}}$
$\Rightarrow \cos (\frac{\pi }{4}+\varphi )=-\frac{la}{n\sqrt{2}}...\left(3\right)$
and
$\Rightarrow \sin (\frac{\pi }{4}+\varphi )=-\frac{mb}{n\sqrt{2}}...\left(4\right)$
Squaring and adding $\left(3\right)$ and $\left(4\right),$ we get
$\frac{l^2{a}^2}{2n^2}+\frac{m^2{b}^2}{2n^2}=1$
$\Rightarrow l^2{a}^2+m^2{b}^2=2n^2$
$\Rightarrow \frac{a^2{l}^2+b^2{m}^2}{n^2}=2$