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.

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.
