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. 

Question: A  collection  of  parallel    chords  connect pairs   of  points  on  an  ellipse,    as  shown

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.



What I have tried/attempted:
Say that a line in the form $y=mx+c$ intersects the ellipse
so 
$$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$
$$ \Leftrightarrow x^2b^2 + y^2a^2 = a^2b^2 $$
$$ \Leftrightarrow x^2b^2 + (mx+c)^2a^2 = a^2b^2 $$
$$\Leftrightarrow x^2b^2 + (m^2x^2+2mcx+c^2)a^2=a^2b^2 $$
$$ \Leftrightarrow x^2b^2 + a^2m^2x^2+2mca^2x+c^2a^2 - a^2b^2 = 0 $$
$$ \Leftrightarrow (b^2+a^2m^2)x^2 + (2mca^2)x + (c^2a^2-a^2b^2) = 0 $$
$$\Leftrightarrow (b^2+a^2m^2)x^2 + (2mca^2)x + a^2(c^2-b^2) = 0 $$
Now I am stuck should I be doing something with this equation I have formed? Or is there a more easy geometrical approach to this question I am missing.
 A: 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. :) 
A: Continuing the computation: the $x$-value of the midpoint is 
$$x_{mid} = \frac{x_1+x_2}{2} = -\frac{mca^2}{b^2+a^2m^2}$$
and the corresponding $y$-value is 
$$y_{mid} = mx_{mid}+c = \frac{b^2c}{b^2+a^2m^2}. $$ 
Therefore, 
$$ma^2 y_{mid} + b^2 x_{mid} = 0$$
and that is an equation of a straight line through the origin.
Remark: that line is the polar of the point at infinity corresponding to the fixed direction with respect to the given conic.
A: WLOG, any ellipse can be transformed to 'fit' the equation $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
And suppose the equation of the chord is $x=k$, $(-a<k<a)$.
There will be two points that the chord and the ellipse meet. Suppose the $y$-coordinate of those two points are $y_1, y_2$. (Then the two points are $(k, y_1)$, $(k, y_2)$)
Then $y_1, y_2$ are solutions of the equation $$\frac{k^2}{a^2}+\frac{y^2}{b^2}=1$$
The problem wants us to find the midpoint of the two intersection, so we want the midpoint, $(\frac{k+k}{2}, \frac{y_1+y_2}{2})=(k, \frac{y_1+y_2}{2})$
But as you can see, the sum of the solutions of $\frac{k^2}{a^2}+\frac{y^2}{b^2}=1$ is $0$.
Thus for any values of $k$, the midpoint of intersections should be on the line $y=0$.
A: $$\Leftrightarrow (b^2+a^2m^2)x^2 + (2mca^2)x + a^2(c^2-b^2) = 0 $$
By the midpoint of a parabola, the $x$-value of the midpoint of the intersection of the line and the ellipse will be at
$$\frac{-2mca^2}{2(b^2+a^2m^2)} = \frac{-mca^2}{b^2+a^2m^2}$$
Note that the $y$ value of the midpoint is linearly dependent on the $x$ value of the midpoint $(y=mx+c)$. Therefore, this expression shows that for any fixed values of $a$, $b$, and $m$, the collection of midpoints for ranging $c$ values is linearly dependent on the variable $c$.
For a concrete example, take the ellipse $\frac{x^2}{4} +\frac{y^2}{9}=1$ intersected by lines with slope $m=3$. The $x$ coordinate for the midpoint of the chords will be
$$x_c=\frac{-mca^2}{b^2+a^2m^2}=\frac{-3\cdot c\cdot4}{9+4\cdot 9}=\frac{-4c}{15}$$
and we can find the corresponding $y$ coordinates of the midpoint by plugging it into our linear equation
$$y_c=mx+c=3\cdot \frac{-4c}{15}+c=\frac{1}{5}c$$
So we have the collection of midpoints $(\frac{-4c}{15},\frac{1}{5}c)$ which is obvious to show lies on a straight line for changing values of $c$.
