Maximum Distance Between Midpoints of Chords in an Ellipse

To find the maximum possible value of the distance between the midpoints $M(P, Q)$ and $M(P, Q')$, we can analyze the position vectors of the points on the ellipse $E: \frac{x^2}{16} + \frac{y^2}{9} = 1$.

Let the position vectors of the distinct points $P, Q,$ and $Q'$ on $E$ be $\vec{P}, \vec{Q},$ and $\vec{Q'}$ respectively.

The midpoint of the line segment joining $P$ and $Q$ is given by: $M(P, Q) = \frac{\vec{P} + \vec{Q}}{2}$

Similarly, the midpoint of the line segment joining $P$ and $Q'$ is given by: $M(P, Q') = \frac{\vec{P} + \vec{Q'}}{2}$

The distance $d$ between the midpoints $M(P, Q)$ and $M(P, Q')$ is: $d = \|M(P, Q) - M(P, Q')\| = \left\| \frac{\vec{P} + \vec{Q}}{2} - \frac{\vec{P} + \vec{Q'}}{2} \right\| = \frac{1}{2} \|\vec{Q} - \vec{Q'}\|$

Thus, the distance between the two midpoints is equal to half the length of the line segment $QQ'$: $d = \frac{1}{2} QQ'$

To maximize $d$, we need to maximize the distance $QQ'$ between two points $Q$ and $Q'$ on the ellipse $E$.

For the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ with semi-major axis $a = 4$ and semi-minor axis $b = 3$, the maximum possible distance between any two points on the ellipse is equal to the length of its major axis: $(QQ')_{\text{max}} = 2a = 2 \times 4 = 8$

This maximum distance is achieved when $Q$ and $Q'$ are the opposite endpoints of the major axis, i.e., $Q = (-4, 0)$ and $Q' = (4, 0)$.

Since $P, Q,$ and $Q'$ must be three distinct points, we can choose $P$ to be any other point on the ellipse not collinear with the major axis, such as $P = (0, 3)$.

Therefore, the maximum possible distance between $M(P, Q)$ and $M(P, Q')$ is: $d_{\text{max}} = \frac{1}{2} (QQ')_{\text{max}} = \frac{1}{2} \times 8 = 4$