Given an ellipse center O, take two perpendicular diameters AOB and COD. Take the diameter A'OB' parallel to the tangents to the ellipse at A and B (this is said to be *conjugate* to the diameter AOB). Similarly, take C'OD' conjugate to COD. Prove that the rectangular hyperbola through A'B'C'D' passes through the foci of the ellipse.

**Solution**

Take the ellipse as x^{2}/a^{2} + y^{2}/b^{2} = 1 and the point A as (a cos t, b sin t). Then AB has slope b/a tan t, so CD has slope -a/b cot t. The tangent at A has slope -b/a cot t. Suppose C is (a cos u, b sin u), then b/a tan u = -a/b cot t and the tangent at C has slope -b/a cot u = b^{3}/a^{3} tan t. Hence the line pair A'B', C'D' has equation (y + x b/a cot t)(y - x b^{3}/a^{3} tan t). Now we have the equations for two distinct conics through A'B'C'D': the original ellipse and the line pair A'B',C'D'. The equation of any other conic through these four points must be a linear combination of the equations of these two, in other words, (x^{2}/a^{2} + y^{2}/b^{2} - 1) + k(y + x b/a cot t)(y - x b^{3}/a^{3} tan t) = 0 for some k.

The criterion for a rectangular hyperbola is that the coefficients of x^{2} and y^{2} should have sum zero, or that 1/a^{2} + 1/b^{2} + k - k b^{4}/a^{4} = 0. Hence k = a^{2}/(b^{4}-b^{2}a^{2}) and the equation of the rectangular hyperbola is x^{2} - y^{2} + (b/a tan t - a/b cot t) xy = a^{2} - b^{2}. But the foci are at (±(a^{2} - b^{2})^{1/2}, 0), so they lie on the rectangular hyperbola.

© John Scholes

jscholes@kalva.demon.co.uk

5 Mar 2002