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 rectum of the given ellipse at the points

• A. $(\pm \frac{3\sqrt{5}}{2},\pm \frac{2}{7})$
• B. $(\pm \frac{3\sqrt{5}}{2},\pm \frac{\sqrt{1}9}{4})$
• C. $(\pm 2\sqrt{3},\pm \frac{1}{7})$
• D. $(\pm 2\sqrt{3},\pm \frac{4\sqrt{3}}{7})$

Answer 1

Answer: (C)

Given, $\frac{x^2}{16}+\frac{y^2}{4}=1$
Here,
$a=4,b=2$
Equation of normal
$4x\sec \theta -2y\mathrm{cosec}\theta =12$
$M\left(\frac{7\cos \theta }{2},\sin \theta \right)=(h,k)$ [say]
$\therefore h=\frac{7\cos \theta }{2}\Rightarrow =\frac{2h}{7}=\cos \theta ....$ (i)
and $k=\sin \theta ....$ (ii)
On squaring and adding Eqs. (i) and (ii), we get
$\frac{4h^2}{49}+k^2=1\left[\because {\cos }^2\theta +{\sin }^2\theta =1\right]$
Hence, locus is $\frac{4x^2}{49}+y^2=1$ .. (iii)

For given ellipse, $e^2=1-\frac{4}{16}=\frac{3}{4}$
$\therefore e=\frac{\sqrt{3}}{2}$
$\therefore x=\pm 4\times \frac{\sqrt{3}}{2}=\pm 2\sqrt{3}[\because x=\pm ae]\dots$ (iv)
On solving Eqs. (iii) and (iv), we get
$\frac{4}{49}\times 12+y^2=1$
$\Rightarrow y^2=1-\frac{48}{49}=\frac{1}{49}$
$y=\pm \frac{1}{7}$
$\therefore$ Required points $\left(\pm 2\sqrt{3},\pm \frac{1}{7}\right)$.