(d) Let mid-point of part $PQ$ which is in between the axis is $R({x_1},{y_1})$ then co-ordinates of $P$ and $Q$ will be $(2{x_1},0)$ and $(0,\,2{y_1})$ respectively.

.$\therefore$  eq. of line $PQ$ is $\frac{x}{{2{x_1}}} + \frac{y}{{2{y_1}}} = 1$ or $y = – \left( {\frac{{{y_1}}}{{{x_1}}}} \right)\,x + 2{y_1}$

If this line touches the ellipse $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1,$

then it will satisfy the condition, ${c^2} = {a^2}{m^2} + {b^2}$

i.e., ${(2{y_1})^2} = {a^2}{\left( {\frac{{ – {y_1}}}{{{x_1}}}} \right)^2} + {b^2}$ or $4y_1^2 = \left\{ {\frac{{{a^2}y_1^2}}{{x_1^2}}} \right\} + {b^2}$

or $4 = \left( {\frac{{{a^2}}}{{x_1^2}}} \right) + \left( {\frac{{{b^2}}}{{y_1^2}}} \right)$or $\left( {\frac{{{a^2}}}{{x_1^2}}} \right) + \left( {\frac{{{b^2}}}{{y_1^2}}} \right) = 4$

$\therefore$  Required locus of $({x_1},{y_1})$ is $\left( {\frac{{{a^2}}}{{{x^2}}} + \frac{{{b^2}}}{{{y^2}}}} \right)$ $=4$.