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/
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