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 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

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

Normal to the ellipse at any point ${}^{\prime }{\varphi }^{\prime }$ is, $4x\sec \varphi -2y\cos ec\varphi =12$

$Q\equiv (3\cos \varphi ,0)$

Let $M\equiv (\alpha ,\beta )$

$\therefore \alpha =\frac{3\cos \varphi +4\cos \varphi }{2}=\frac{7}{2}\cos \varphi$

$\Rightarrow \cos \varphi =\frac{2}{7}\alpha$
and $\beta =\sin \varphi$

Now using, ${\cos }^2\varphi +{\sin }^2\varphi =1$

$\Rightarrow \frac{4}{49}{\alpha }^2+{\beta }^2=1\Rightarrow \frac{4}{49}{x}^2+y^2=1$
$\Rightarrow$ latus rectum $x=\pm 2\sqrt{3}$
$\frac{48}{49}+y^2=1\Rightarrow y=\pm \frac{1}{7}$
Thus required point is $(\pm 2\sqrt{3},\pm \frac{1}{7})$