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·d^{2} is independent of the position of P.

**Solution**

Answer: a^{2}b^{2}.

Let the ellipse be x^{2}/a^{2} + y^{2}b^{2} = 1. Then the foci are at (± ae, 0) where the eccentricity e is given by b^{2} = a^{2}(1 - e^{2}). It is also a well-known property that PA + PB = 2a. Thus we may express the product 2 PA·PB as (PA + PB)^{2} - PA^{2} - PB^{2} = 4a^{2} - (x + ae)^{2} - y^{2} - (x - ae)^{2} - y^{2} = 4a^{2} - 2x^{2} - 2y^{2} - 2a^{2}e^{2} = 2a^{2} + 2b^{2} - 2x^{2} - 2y^{2}.

The tangent at (x, y) is x/b^{2} Y + x/a^{2} X = 1, so it meets the axes at (a^{2}/x, 0), (0, b^{2}/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 a^{2}b^{2}/(xy) or as 1/2 d times hypoteneuse. Hence d^{2} = a^{4}b^{4}/( (xy)^{2} (a^{4}x^{2} + b^{4}/y^{2}) = a^{4}b^{4}/(a^{4}y^{2} + b^{4}x^{2}). Using the fact that (x, y) lies on the ellipse, we have a^{4}y^{2} = a^{4}b^{2} - a^{2}b^{2}x^{2}, and b^{4}x^{2} = a^{2}b^{4} - a^{2}b^{2}y^{2}, so a^{4}y^{2} + b^{4}x^{2} = a^{2}b^{2} (a^{2} + b^{2} - x^{2} - y^{2}) = a^{2}b^{2}PA·PB. Hence d^{2} = a^{2}b^{2}/( PA·PB ).

© John Scholes

jscholes@kalva.demon.co.uk

23 Jan 2001