MCQ
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
    $a^2l^2 + b^2n^2 = 2 m^2$
  • B
    $a^2m^2 + b^2l^2 = 2 n^2$
  • $a^2l^2 + b^2m^2 = 2 n^2$
  • D
    $a^2n^2 + b^2m^2 = 2 l^2$

Answer

Correct option: C.
$a^2l^2 + b^2m^2 = 2 n^2$
c
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 $  

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