A particle moves in the plane so that its angular velocity about the point (1, 0) equals minus its angular velocity about the point (-1, 0). Show that its trajectory satisfies the differential equation y' x(x^{2} + y^{2} - 1) = y(x^{2} + y^{2} + 1). Verify that this has as solutions the rectangular hyperbolae with center at the origin and passing through (±1, 0).

**Solution**

The angular momentum (for unit mass) about (1, 0) is (x - 1) dy/dt - y dx/dt. Hence the angular velocity is ( (x - 1) dy/dt - y dx/dt )/( (x - 1)^{2} + y^{2}). Similarly, the angular velocity about (-1, 0) is ( (x + 1) dy/dt - y dx/dt )/( (x + 1)^{2} + y^{2}). Hence the trajectory satisfies: ( (x - 1) y' - y)(x^{2} + y^{2} + 1 + 2x) + ( (x + 1) y' - y)(x^{2} + y^{2} + 1 - 2x) = 0 or 2x y' (x^{2} + y^{2} + 1) - 4x y' - 2y(x^{2} + y^{2} + 1) = 0, or x y' (x^{2} + y^{2} - 1) = y (x^{2} + y^{2} + 1).

We see immediately that xy = 0 is a solution and almost immediately that x^{2} - y^{2} = 1 is a solution. Hence also all linear combinations x^{2} + k xy - y^{2} = 1. These are rectancular hyperbolae because the sum of the coefficients of x^{2} and y^{2} are zero.

© John Scholes

jscholes@kalva.demon.co.uk

5 Mar 2002