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If $l x+m y+n=0$ represents a chord of the ellipse $b^2 x^2+a^2 y^2=a^2 b^2$ whose eccentric angles differ by $90^{\circ}$, then
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Verified Answer
The correct answer is:
$a^2 l^2+b^2 m^2=2 n^2$
Equation of chord joining points $P(a \cos \alpha, b \sin \alpha)$ and $Q(a \cos \beta, b \sin \beta)$ is $\frac{x}{a} \cos \left(\frac{\alpha+\beta}{2}\right)+\frac{y}{b} \sin \left(\frac{\alpha+\beta}{2}\right)=\cos \left(\frac{\alpha-\beta}{2}\right)$
Now, $\beta=\alpha+90^{\circ}$
$\frac{x}{a} \cos \left(\frac{2 \alpha+90^{\circ}}{2}\right)+\frac{y}{b} \sin \left(\frac{2 \alpha+90^{\circ}}{2}\right)=\frac{1}{\sqrt{2}}$
now, compare it with, $l x+m y=-n$,
we get
$\begin{aligned}
& \frac{\cos \left(\frac{2 a n a 0^{\circ}}{2}\right)}{a l}=\frac{\sin \left(\frac{2 \alpha a 0^{\circ}}{20^{\circ}}\right)}{b m}=-\frac{1}{\sqrt{2} n} \\
& \because \cos ^2 \theta+\sin ^2 \theta=1 \\
& \Rightarrow a^2 l^2+b^2 m^2=2 n^2
\end{aligned}$
Now, $\beta=\alpha+90^{\circ}$
$\frac{x}{a} \cos \left(\frac{2 \alpha+90^{\circ}}{2}\right)+\frac{y}{b} \sin \left(\frac{2 \alpha+90^{\circ}}{2}\right)=\frac{1}{\sqrt{2}}$
now, compare it with, $l x+m y=-n$,
we get
$\begin{aligned}
& \frac{\cos \left(\frac{2 a n a 0^{\circ}}{2}\right)}{a l}=\frac{\sin \left(\frac{2 \alpha a 0^{\circ}}{20^{\circ}}\right)}{b m}=-\frac{1}{\sqrt{2} n} \\
& \because \cos ^2 \theta+\sin ^2 \theta=1 \\
& \Rightarrow a^2 l^2+b^2 m^2=2 n^2
\end{aligned}$
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