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.

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 $y=0$.
