Let an ellipse have center O and foci A and B. For a point P on the ellipse let d be the distance from O to the tangent at P. Show that PA·PB·d2 is independent of the position of P.
Solution
Answer: a2b2.
Let the ellipse be x2/a2 + y2b2 = 1. Then the foci are at (± ae, 0) where the eccentricity e is given by b2 = a2(1 - e2). It is also a well-known property that PA + PB = 2a. Thus we may express the product 2 PA·PB as (PA + PB)2 - PA2 - PB2 = 4a2 - (x + ae)2 - y2 - (x - ae)2 - y2 = 4a2 - 2x2 - 2y2 - 2a2e2 = 2a2 + 2b2 - 2x2 - 2y2.
The tangent at (x, y) is x/b2 Y + x/a2 X = 1, so it meets the axes at (a2/x, 0), (0, b2/y). These two points and the origin form a right-angled triangle with the origin a height d from the hypoteneuse. So we may calculate its area as 1/2 a2b2/(xy) or as 1/2 d times hypoteneuse. Hence d2 = a4b4/( (xy)2 (a4x2 + b4/y2) = a4b4/(a4y2 + b4x2). Using the fact that (x, y) lies on the ellipse, we have a4y2 = a4b2 - a2b2x2, and b4x2 = a2b4 - a2b2y2, so a4y2 + b4x2 = a2b2 (a2 + b2 - x2 - y2) = a2b2PA·PB. Hence d2 = a2b2/( PA·PB ).
© John Scholes
jscholes@kalva.demon.co.uk
23 Jan 2001