Let E be the ellipse $$\frac{x^{2}}{16} + \frac{y^{2}}{9} = 1$$. For any three distinct points 𝑃,Q and $$Q^{'}$$ on E, let M(P, Q) be the mid-point of the line segment joining P and Q, and $$M(P, Q^{'})$$ be the mid-point of the line segment joining P and $$Q^{'}$$. Then the maximum possible valu of the distance between M(P, Q) and $$M(P, Q^{'})$$ as P, Q and $$Q^{'}$$ vary on E, is

The ellipse is $$\frac{x^{2}}{16}+\frac{y^{2}}{9}=1$$, so its semi-major axis is $$a=4$$ and its semi-minor axis is $$b=3$$.
Hence the length of the major axis is $$2a = 8$$.

Choose any point $$P(x_{1},y_{1})$$ on the ellipse. Let $$Q(x_{2},y_{2})$$ and $$Q'(x_{3},y_{3})$$ be two other (distinct) points on the same ellipse.

The mid-point of the segment $$PQ$$ is $$M(P,Q)=\left(\frac{x_{1}+x_{2}}{2},\ \frac{y_{1}+y_{2}}{2}\right).$$

The mid-point of the segment $$PQ'$$ is $$M(P,Q')=\left(\frac{x_{1}+x_{3}}{2},\ \frac{y_{1}+y_{3}}{2}\right).$$

Distance between these two mid-points:

$$\begin{aligned} d &= \sqrt{\left(\frac{x_{1}+x_{2}}{2}-\frac{x_{1}+x_{3}}{2}\right)^{2}+ \left(\frac{y_{1}+y_{2}}{2}-\frac{y_{1}+y_{3}}{2}\right)^{2}}\\[4pt] &= \sqrt{\left(\frac{x_{2}-x_{3}}{2}\right)^{2}+ \left(\frac{y_{2}-y_{3}}{2}\right)^{2}}\\[4pt] &=\frac{1}{2}\,\sqrt{(x_{2}-x_{3})^{2}+(y_{2}-y_{3})^{2}}\\[4pt] &=\frac{1}{2}\;\text{(distance between }Q\text{ and }Q'). \end{aligned}$$

Therefore maximising $$d$$ is the same as maximising the distance between $$Q$$ and $$Q'$$ on the ellipse.

The greatest possible distance between two points on an ellipse is the length of its major axis, namely $$8$$ (achieved at the end-points $$(4,0)$$ and $$(-4,0)$$).

Hence the maximum value of $$d$$ is $$d_{\text{max}}=\frac{1}{2}\times 8 = 4.$$

Answer: 4