For the general ellipse $ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ ,show that the midpoints of the chords lie on a straight line.

Apply the linear transformation $(x, y) \mapsto (ax, by)$. The transforms your ellipse to the unit circle. It also transforms parallel lines to parallel lines, and midpoints to midpoints. Thus if you can prove the statement for the unit circle, you're done. But for the unit circle, the chords orthogonal to the vector $(p, q)$ have midpoints on the line $-qx + py = 0$. QED.

Added detail:

A vector $(x, y)$ is orthogonal to the vector $(p, q)$ if and only iff their dot product is zero. That means that the vector $(-q, p)$ is orthogonal to $(p, q)$. So lines orthogonal to $(p, q)$ all have the form $$ \{ (a, b) + t (-q, p) \mid t \in \mathbb R \} $$ where $(a, b)$ can be any point. So let's pick a point that happens to be on the line through the origin in direction $(p, q)$, i.e., $$ (a, b) = s(p, q) = (sp, sq). $$ Now for a point of that line to lie on the unit circle, we need $$ \| (sp, sq) + t(-q, p) \|= 1. $$ That means that $$ \| (sp, sq) + t(-q, p) \|^2= 1, $$ which can be written $$ \left( (sp, sq) + t(-q, p) \right) \cdot \left( (sp, sq) + t(-q, p) \right) = 1. $$ Since the $(sp, sq)$ and $(-q, p)$ are perpendicular, this expands to just $$ (sp, sq) \cdot (sp, sq) + t^2 (-q, p) \cdot (-q, p) = 1 $$ or $$ (s^2 + t^2) K= 1 $$ where $K = p^2 + q^2$. If $t$ is a solution to this equation, then so is $-t$, so the two points of the chord correspond to $\pm t$, and their average corresponds to $t = 0$, i.e, to the point $(x, y) = (sp, sq)$, which happens to be on the line perpendicular to $(-q, p)$. So for a chord-center $(x, y)$, we have $(-q, p) \cdot (x, y) = 0$, which becomes $$ -qx + py = 0. $$

One last thought: Perhaps a better way to go is to first do the transform, and then observe that the resulting situation is circularly symmetric, so we might as well choose, as our set of chords, the vertical chords. Their midpoints obviously lie on the $x$-axis, and we're done.

To some degree, that's what my discussion above did: the vectors $(p, q)$ and $(-q, p)$ are basis vectors for a rotated frame of reference in which the chords are vertical. :)