Explanation

(a)
Let the point be $P(\sqrt{2}\cos{\theta,\sin{\theta)}on\ \frac{x^{2}}{2} + \frac{y^{2}}{1} = 1}$
$\Rightarrow$ Equation of tangent is
$\frac{x\sqrt{2}}{2}\cos{\theta + y\sin\theta = 1}$
Whose intercept on coordinate axes are $A(\sqrt{2}\sec\theta,\ 0)$ and $B(0,\ cosec\ \theta\ )$
$\therefore$ Mid point of its intercept between axes is
$\left( \frac{\sqrt{2}}{2}\sec\theta,\frac{1}{2}cosec\ \theta \right) = (h,\ k)$
$\cos{\theta = \frac{1}{\sqrt{2h}}}\ and\sin{\theta = \frac{1}{2k}}$
Thus, locus of mid point $M$ is
$\cos^{2}{\theta + \sin^{2}{\theta = \frac{1}{2h^{2}} + \frac{1}{4k^{2}}}}$
$\Rightarrow \ \frac{1}{2x^{2}} + \frac{1}{4y^{2}} = 1$