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 latusrectum of the given ellipse at the points

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

$(\pm 2\sqrt{3},\pm \frac{1}{7})$

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

For given ellipse, $e^2=1-\frac{4}{16}=\frac{3}{4}$
$\therefore e=\frac{\sqrt{3}}{2}$
$\therefore x=\pm 4\times \frac{\sqrt{3}}{2}=\pm 2\sqrt{3}[\because x=\pm ae]$ ... (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 $(\pm 2\sqrt{3},\pm \frac{1}{7})$.