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The line, $ lx + my + n = 0$ will cut the ellipse $\frac{{{x^2}}}{{{a^2}}}$ $+$ $\frac{{{y^2}}}{{{b^2}}}$ $= 1 $ in points whose eccentric angles differ by $\pi /2$ if :
$a^2l^2 + b^2n^2 = 2 m^2$
$a^2m^2 + b^2l^2 = 2 n^2$
$a^2l^2 + b^2m^2 = 2 n^2$
$a^2n^2 + b^2m^2 = 2 l^2$
Solution
Equation of a chord
$\frac{x}{a} \cos \frac{{\alpha \,\, + \,\,\beta }}{2} + \frac{y}{b} \sin \frac{{\alpha \,\, + \,\,\beta }}{2} = cos \frac{{\alpha \,\, – \,\,\beta }}{2}$
Put $ \beta = \alpha + \frac{\pi }{2}$ , equation reduces to,
$bx (cos \alpha – sin \alpha ) + ay (cos \alpha + sin \alpha ) = ab$…$(1)$
compare with $l x + my = – n$…..$(2) $
$\left. {\begin{array}{*{20}{c}} {\cos \,\alpha \,\, – \,\,\sin \,\alpha \,\,\, = \,\,\,{\textstyle{{a\,\ell } \over { – \,n}}}}\\ {\cos \,\alpha \,\, + \,\,\sin \,\alpha \,\,\, = \,\,\,{\textstyle{{m\,b} \over { – \,n}}}} \end{array}} \right\}$Squaring and adding $a^2 l^2 + b^2 m^2 – 2 n^2 = 0 $
