Question

The normal at a point $P$ on the ellipse $x^2+4y^2=16$ intersects 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 points

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

$x^2+4y^2=16$
$\Rightarrow \frac{x^2}{16}+\frac{y^2}{4}=1;a^2=16,b^2=4$


$P(4\cos \theta ,2\sin \theta )$
Equation of tangent at $P:$
$\frac{x\cos \theta }{4}+\frac{y\sin \theta }{2}=1$

Slope of tangent, $m_t=\frac{-1}{2}\cot \theta$
$\therefore$ Slope of normal, $m_n=2\tan \theta$

Equation of normal at $P:$
$y-2\sin \theta =2\tan \theta (x-4\cos \theta )$
Normal intersects the $x$-axis.
$\Rightarrow y=0$
$-2\sin \theta =\frac{2\sin \theta }{\cos \theta }(x-4\cos \theta )$ $\Rightarrow x=3\cos \theta$
Therefore, coordinates of $Q$ is $(3\cos \theta ,0)$

$P(4\cos \theta ,2\sin \theta );Q(3\cos \theta ,0)$
$M:(\frac{7}{2}\cos \theta ,\sin \theta )$
Let $M$ be $(h,k)$
${\left(\frac{2}{7}h\right)}^2+k^2=1$
$\Rightarrow \frac{x^2}{{\left(\frac{7}{2}\right)}^2}+\frac{y^2}{1}=1$, which represents an ellipse.

Equation of latus rectum : $x=\pm ae$
$e^2=1-\frac{b^2}{a^2}=1-\frac{4}{16}=\frac{12}{16}$
$\Rightarrow e=\frac{\sqrt{3}}{2}$
$x=\pm 2\sqrt{3}$

Now,
$\frac{x^2}{{\left(\frac{7}{2}\right)}^2}+\frac{y^2}{1}=1;x=\pm 2\sqrt{3}$
$\Rightarrow y=\pm \frac{1}{7}$
Points of intersection $(\pm 2\sqrt{3},\pm \frac{1}{7})$