Question

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

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

Answer 1

The correct option is C $(\pm 2\sqrt{3},\pm \frac{1}{7})$

Given ellipse is
$\frac{x^2}{16}+\frac{y^2}{4}=1\dots \left(1\right)$
Let point $P$ is $(4\cos \varphi ,2\sin \varphi )$
Normal at point $P$ is given by $4x\sec \varphi -2y\text{cosec}\varphi =12$


Normal intersects $x-$ axis at $Q\equiv (3\cos \varphi ,0)$

Let mid point of $PQ$ is $M\equiv (\alpha ,\beta ),$ then
$\Rightarrow \alpha =\frac{3\cos \varphi +4\cos \varphi }{2}=\frac{7}{2}\cos \varphi$
$\Rightarrow \cos \varphi =\frac{2}{7}\alpha$
and $\beta =\sin \varphi$

Using ${\cos }^2\varphi +{\sin }^2\varphi =1,$ we have
$\frac{4}{49}{\alpha }^2+{\beta }^2=1$
$\Rightarrow \frac{4}{49}{x}^2+y^2=1\dots \left(2\right)$

Now, equation of latus rectum of ellipse $\left(1\right)$ is
$x=\pm ae$
$\Rightarrow x=\pm 2\sqrt{3}\dots \left(3\right)$

Solving $\left(2\right)$ and $\left(3\right)$, we get
$\frac{48}{49}+y^2=1$
$\Rightarrow y=\pm \frac{1}{7}$
Hence points of intersection are $(\pm 2\sqrt{3},\pm \frac{1}{7})$