4th Putnam 1941

Problem B4

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.



Take the ellipse as x2/a2 + y2/b2 = 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 = b3/a3 tan t. Hence the line pair A'B', C'D' has equation (y + x b/a cot t)(y - x b3/a3 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, (x2/a2 + y2/b2 - 1) + k(y + x b/a cot t)(y - x b3/a3 tan t) = 0 for some k.

The criterion for a rectangular hyperbola is that the coefficients of x2 and y2 should have sum zero, or that 1/a2 + 1/b2 + k - k b4/a4 = 0. Hence k = a2/(b4-b2a2) and the equation of the rectangular hyperbola is x2 - y2 + (b/a tan t - a/b cot t) xy = a2 - b2. But the foci are at (±(a2 - b2)1/2, 0), so they lie on the rectangular hyperbola.



4th Putnam 1941

© John Scholes
5 Mar 2002