The set of midpoints of the parallel chords of an ellipse with gradient, constant 'm', lie on a straight line: find its equation; equation of ellipse: x^2 + 4y^2 = 4

// Allow me to make the question more explicit: we have a set of lines of equal gradients ‘m’. We need to find a line through which the midpoints of the intersections of these lines with the ellipse; an ellipse is symmetrical, so we can conjecture that the y-intercept will be zero – for this is the centre of the ellipse, and thus our result should be an ‘x’ with a constant co-efficient only.

A chord will connect two points across the ellipse. If we were to draw a line across it, we know that the midpoint of the intersection of a line – regardless of its equation, will be the sum of the co-ordinates, divided by 2 – i.e. the mean co-ordinate. So, let’s do that:

Take the general equation of a line:

y = mx + c, and see where the co-ordinates of the intersection we get when the variables are equal, so we make the substitution into the equation of the ellipse:

   1)    y = mx + c

    2)   x² + 4y² = 4

   =>   x²+ 4(mx+x)² = 4

Expanding:

=>   x²+ 4( [mx]² +  2mxc  +  c² )  =  4

Rearranging:

=>   {1 + 4m²} x²  +  {8mc}x  +  (4c² - 4)  =  0

This is a quadratic – the mean of the roots is when the there minimum is equal to zero.

Differentiating and setting equal to zero:

=>   2{1 + 4m²}x  +  8mc  =  0

Knowing this is satisfied for our chords to go through the midpoints, rearrange for the intercept ‘c’:

=>    c =  - {1 + 4m²}x / 4m

Observe that when x = 0, c = 0, which is what we predicted from the symmetry of an ellipse.

We now substitute this into the equation of the line:

y = mx + c

   = mx – {1 + 4m^2}x / 4m

   = {4m^2 – 1 – 4m^2}x / 4m

   = -x/(4m)

This is our result. Isn't that nice? //

You can generalise this result for the general equation of an ellipse - perhaps a good exercise, and, as a challenge, you can try to solve for a translocated ellipse [even rotated through an arbitrary angle if you want more of a challenge!].