Explanation

(c)
Given, $\frac{x^{2}}{16} + \frac{y^{2}}{4} = 1$
Here, $a = 4,\ b = 2$
Equation of normal
$4x\sec\theta - 2\ y\ cosec\ \theta = 12$
Since, it passes through$\ \mathcal{Q}$ on x-axis
So, put $y = 0$, we get
$x = 3\cos\theta$
$\therefore\ \mathcal{Q}(3\cos{\theta,\ 0)}$
Now, mid point of $P\mathcal{Q}$,
$M\left( \frac{7\cos\theta}{2},\sin\theta \right) = (h,\ k)$ (say)
$\therefore h = \frac{7\cos\theta}{2}\ \Rightarrow \ \cos{\theta = \frac{2h}{7}}$
and $k = \sin\theta$
$\Rightarrow \ \frac{4h^{2}}{49} = k^{2} = 1\ \ (\because\cos^{2}\theta\ + \sin^{2}\theta = 1)$
Here, locus of $M$ is $\frac{4x^{2}}{49} + y^{2} = 1$ …(i)
[image:11_image72.emf]
For given ellipse $e^{2} = 1 - \frac{4}{16} = \frac{3}{4}$
$\therefore\ e = \frac{\sqrt{3}}{2}$
$\therefore$ Abscissa of focus is
$x = \pm 4 \times \frac{\sqrt{3}}{2} = \pm 2\sqrt{3}\ \ \ \ (\because x = \pm ae)$ …(iii)
On solving Eqs. (i) and (ii), we get
$\frac{4}{49} \times 12 + y^{2} = 1$
$\Rightarrow \ y^{2} = 1 - \frac{48}{49} = \frac{1}{49}$
$\therefore$ Required points$\ \left( \pm 2\sqrt{3},\ \pm \frac{1}{7} \right)$