Show that the solutions of the differential equation (9 - x2) (y')2 = 9 - y2 are conics touching the sides of a square.
We have dy/dx = ± √(1 - (y/3)2) / √(1 - (x/3)2. Integrating, sin-1y/3 = ± sin-1x/3 + A, where A is a constant. Hence y/3 = sin A cos(±sin-1x/3) ± x/3 cos A = sin A √(1 - (x/3)2) ± x/3 cos A. So y = k (9 - x2) ± h x, where h2 + k2 = 1. So y2 ± 2hxy + x2 = 9(1 - h2). This is evidently a family of conics. Note that h is not restricted to being less than 1. By taking a suitable complex constant A we can get any value for h. But since negative values are possible, there is no benefit in retaining the ± sign. So we can write the family of conics simply as y2 + 2hxy + x2= 9(1 - h2).
We might guess that the square is x = ±3, y = ±3. If we substitute x = 3 in the equation for the conic, we get y2 + 6hy + 9 = 9 - 9 h2, or (y - 3h)2 = 0. This has a repeated root y = 3h, which shows that the line x = 3 touches the conic. Similarly for the other four sides of the square.
12th Putnam 1952
© John Scholes
5 Mar 2002