(c) Let the point of contact be

$R \equiv (\sqrt 2 \cos \theta ,\,\sin \theta )$

Equation of tangent $AB$ is

$\frac{x}{{\sqrt 2 }}\cos \theta + y\sin \theta = 1$

$ \Rightarrow $ $A \equiv (\sqrt 2 \sec \theta ,\,0);\,B \equiv (0,\,{\rm{cosec }}\theta )$

Let the middle point $Q$ of $AB$ be $(h,\,k)$

$ \Rightarrow $ $h = \frac{{\sec \theta }}{{\sqrt 2 }},\,k = \frac{{{\rm{cosec }}\theta }}{2} $

$\Rightarrow \cos \theta = \frac{1}{{h\sqrt 2 }},\,\sin \theta = \frac{1}{{2k}}$

$ \Rightarrow $ $\frac{1}{{2{h^2}}} + \frac{1}{{4{k^2}}} = 1$,

Required locus is $\frac{1}{{2{x^2}}} + \frac{1}{{4{y^2}}} = 1$.

Trick : The locus of mid-points of the portion of tangents to the ellipse $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$ intercepted between axes is

${a^2}{y^2} + {b^2}{x^2} = 4{x^2}{y^2}$

$i.e.$, $\frac{{{a^2}}}{{4{x^2}}} + \frac{{{b^2}}}{{4{y^2}}} = 1$

or $\frac{1}{{2{x^2}}} + \frac{1}{{4{y^2}}} = 1$.