Explanation

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