Show that the solutions of the differential equation (9 - x^{2}) (y')^{2} = 9 - y^{2} are conics touching the sides of a square.

**Solution**

We have dy/dx = ± √(1 - (y/3)^{2}) / √(1 - (x/3)^{2}. Integrating, sin^{-1}y/3 = ± sin^{-1}x/3 + A, where A is a constant. Hence y/3 = sin A cos(±sin^{-1}x/3) ± x/3 cos A = sin A √(1 - (x/3)^{2}) ± x/3 cos A. So y = k (9 - x^{2}) ± h x, where h^{2} + k^{2} = 1. So y^{2} ± 2hxy + x^{2} = 9(1 - h^{2}). 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 y^{2} + 2hxy + x^{2}= 9(1 - h^{2}).

We might guess that the square is x = ±3, y = ±3. If we substitute x = 3 in the equation for the conic, we get y^{2} + 6hy + 9 = 9 - 9 h^{2}, 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.

© John Scholes

jscholes@kalva.demon.co.uk

5 Mar 2002