Q98 / 200 · Manipal (MET) 2013 · Mathematics

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

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

Solution

Let the points of intersection of the line and the ellipse be $(a \cos \theta, b \sin \theta)$ and $\left\{a \cos \left(\frac{\pi}{2}+\theta\right), b \sin \left(\frac{\pi}{2}+\theta\right)\right\}$, since they lie on the given line $\quad l x+m y+n=0$. $l a \cos \theta+m b \sin \theta+n=0$ $\Rightarrow \quad l a \cos \theta+m b \sin \theta=-n$ and $l a \sin \theta+m b \cos \theta+n=0$ $l a \sin \theta-m b \cos \theta=n$ squarring and adding, we get, $a^2 l^2+b^2 m^2=2 n^2 \quad \Rightarrow \frac{a^2 l^2+b^2 m^2}{n^2}=2$.

Hence, the correct answer is option: 'B'.

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