A Proof using Analytic Geometry

Date: 02/24/99 at 05:42:18
From: Ngu Soon Hui
Subject: Geometry

Prove that, if p is a point inside the ellipse, there is one and only 
one chord QP bisected at P.

Date: 02/24/99 at 16:16:40
From: Doctor Rob
Subject: Re: Geometry

This is false if the point is the center of the ellipse, but true
otherwise. Here is a proof using analytic geometry.

Pick the coordinate system so that the center of the ellipse is the
origin, the x-axis is the major axis of the ellipse, and the y-axis is 
the minor axis of the ellipse.

Then, the equation of the ellipse is x^2/a^2 + y^2/b^2 = 1. Let the
coordinates of the point be (x0,y0). Let a chord through (x0,y0) have 
slope m. Then the equation of the chord is m = (y-y0)/(x-x0). The 
endpoints of the chord (on the ellipse) are

x1 = a*(a*m*[m*x0-y0]-b*Sqrt[b^2+a*m^2-(y0-m*x0)^2])/(b^2+a^2*m^2),
y1 = -b*(b*m*[m*x0-y0]+a*m*Sqrt[b^2+a*m^2-(y0-m*x0)^2])/(b^2+a^2*m^2),

x2 = a*(a*m*[m*x0-y0]+b*Sqrt[b^2+a*m^2-(y0-m*x0)^2])/(b^2+a^2*m^2),
y2 = -b*(b*m*[m*x0-y0]-a*m*Sqrt[b^2+a*m^2-(y0-m*x0)^2])/(b^2+a^2*m^2).

The denominators can never be zero, since both a and b are positive. 
The quantity inside the square root is positive since (x0,y0) is
specified to be inside the ellipse. The midpoint of the chord is

   x3 = a^2*m*(m*x0-y0)/(b^2+a^2*m^2),
   y3 = -b^2*(m*x0-y0)/(b^2+a^2*m^2).

Setting these equal to x0 and y0 and solving for m leads to the 
equation

   b^2*x0 + a^2*m*y0 = 0,

which is linear in m, and has the unique solution

   m = -b^2*x0/(a^2*y0),

unless y0 = 0. If y0 = 0 and x0 = 0, then any value of m works. If
y0 = 0 and x0 is nonzero, then no finite value of m works. To deal with 
m = infinity, when the chord is vertical, its equation is x = x0. The 
midpoint of the chord is (x0,0), which equals (x0,y0). Thus there is 
still a unique chord bisected by P, namely the vertical one.

Thus we have proved that unless P = (0,0), there is a unique chord of
an ellipse bisected by P, and if P = (0,0), then every chord of the
ellipse is bisected by P.

- Doctor Rob, The Math Forum
  http://mathforum.org/dr.math/