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 $P Q$, then the locus of $M$ intersects the latusrectum of the given ellipse at the points

• A

  $\left(\pm \frac{3 \sqrt{5}}{2}, \pm \frac{2}{7}\right)$
• B

  $\left(\pm \frac{3 \sqrt{5}}{2}, \pm \frac{\sqrt{19}}{4}\right)$
• Correct answer
  C

  $\left(\pm 2 \sqrt{3}, \pm \frac{1}{7}\right)$
• D

  $\left(\pm 2 \sqrt{3}, \pm \frac{4 \sqrt{3}}{7}\right)$

Solution

Given, $\quad \frac{x^2}{16}+\frac{y^2}{4}=1$ Here, $\quad a=4, b=2$ Equation of normal is $$ \begin{aligned} & 4 x \sec \theta-2 y \operatorname{cosec} \theta=12 \\ & M\left(\frac{7 \cos \theta}{2}, \sin \theta\right)=(h, k) \text { (say) } \\ & \therefore \quad h=\frac{7 \cos \theta}{2} \Rightarrow \cos \theta=\frac{2 h}{7} \\ & {\left[\because \cos ^2 \theta+\sin ^2 \theta=1\right]} \\ & \end{aligned} $$ Hence, locus is $\frac{4 x^2}{49}+y^2=1$
For given ellipse, $e^2=1-\frac{4}{16}=\frac{3}{4}$ $$ \begin{array}{rc} \therefore \quad e=\frac{\sqrt{3}}{2} \\ \therefore \quad x=\pm 4 \times \frac{\sqrt{3}}{2}=\pm 2 \sqrt{3} \\ & {[\because x=\pm a e] \ldots \text { (ii) }} \end{array} $$ On solving Eqs. (i) and (ii), we get $$ \begin{aligned} & \frac{4}{49} \times 12+y^2=1 \Rightarrow y^2=1-\frac{48}{49}=\frac{1}{49} \\ & \Rightarrow \quad y=\pm \frac{1}{7} \end{aligned} $$ $\therefore$ Required points $\left(\pm 2 \sqrt{3}, \pm \frac{1}{7}\right)$.

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