S is a solid square side 2a. It lies in the quadrant x ≥ 0, y ≥ 0, and it is free to move around provided a vertex remains on the x-axis and an adjacent vertex on the y-axis. P is a point of S. Show that the locus of P is part of a conic. For what P does the locus degenerate?

**Solution**

Let A be the vertex that moves along the x-axis and B the vertex that moves along the y-axis. Suppose that when AB is horizontal P has coordinates b, c. In the general configuration let be the angle BAO be θ. Then P has coordinates x = (2a - b) cos θ + c sin θ, y = b sin θ + c cos θ. Hence cx - (2a - b)y = (b^{2} + c^{2} - 2ab) sin θ, bx - cy = (2ab - b^{2} - c^{2}) cos θ. Squaring and adding we eliminate θ to get: (b^{2} + c^{2}) x^{2} - 4ac xy + (4a^{2} + b^{2} + c^{2} - 4ab) y^{2} = (b^{2} + c^{2} - 2ab)^{2}, which is the equation of a conic. So the locus of P must form part of this conic.

The conic degenerates if b^{2} + c^{2} = 2ab. In this case, the equation becomes 2ab x^{2} - 4ac xy + (4a^{2} - 2ab)y^{2} = 0, or bx^{2} - 2c xy + (2a - b)y^{2} = 0, or b^{2}x^{2} - 2bc xy + c^{2} y^{2} = 0, or bx = cy. So in this case the locus lies on a straight line. We may write the condition b^{2} + c^{2} = 2ab as (a - b)^{2} + c^{2} = a^{2}, which shows that such P lie on the semicircle diameter AB.

© John Scholes

jscholes@kalva.demon.co.uk

5 Mar 2002