Question

The normal at a point $P$ on the ellipse $x^{2}+4 y^{2}=16$ meets the $x$-axis at $Q$. If $M$ is the mid point of the line segment $PQ$, then the locus of $M$ intersects the latus rectums of the given ellipse at the points

• A. $\left(\pm \frac{3 \sqrt{5}}{2}, \pm \frac{2}{7}\right)$
• B. $\left(\pm \frac{3 \sqrt{5}}{2}, \pm \frac{\sqrt{19}}{4}\right)$
• C. $\left(\pm 2 \sqrt{3}, \pm \frac{1}{7}\right)$
• D. $\left(\pm 2 \sqrt{3}, \pm \frac{4 \sqrt{3}}{7}\right)$

Answer 1

Answer: (C)

Normal is $4 x \sec \phi-2 y \operatorname{cosec} \phi=12 $
$Q=(3 \cos \phi, 0) $
$M=(\alpha, \beta) $
$\alpha=\frac{3 \cos \phi+4 \cos \phi}{2}=\frac{7}{2} \cos \phi$
$\Rightarrow \cos \phi=\frac{2}{7} \alpha$
$\beta=\sin \phi $
$\cos ^{2} \phi+\sin ^{2} \phi=1 $
$\Rightarrow \frac{4}{49} \alpha^{2}+\beta^{2}=1$
$ \Rightarrow \frac{4}{49} x^{2}+y^{2}=1$
$\Rightarrow $ latus rectum $ x=\pm 2 \sqrt{3} $
$\frac{48}{49}+y^{2}=1 $
$\Rightarrow y=\pm \frac{1}{7} $
$(\pm 2 \sqrt{3}, \pm 1 / 7)$.