Question

If the line $lx+my+n=0$ cuts the ellipse $\frac{x^2}{a^2}+\frac{y^2}{25}=1$ in points whose eccentric angles differ by $\frac{\pi }{2}$, then $\frac{a^2{l}^2+b^2{m}^2}{n^2}$ is equal to

• A. $1$
• B. $2$
• C. $4$
• D. $\frac{3}{2}$

Answer 1

The correct option is B $2$

Let the points of intersection of the line and the ellipse be $(a\cos \theta ,b\sin \theta )$ and $\{a\cos (\frac{\pi }{2}+\theta ),b\sin (\frac{\pi }{2}+\theta )\}$

Since they lie on the given line $lx+my+n=0$.
$la\cos \theta +mb\sin \theta +n=0$
$\Rightarrow la\cos \theta +mb\sin \theta =-n$
and $la\sin \theta +mb\cos \theta +n=0$
$la\sin \theta -mb\cos \theta =n$
On squaring and adding, we get,
$a^2{l}^2+b^2{m}^2=2n^2$

$\Rightarrow \frac{a^2{l}^2+b^2{m}^2}{n^2}=2$