Find the surface comprising the curves which satisfy dx/(yz) = dy/(zx) = dz/(xy) and which meet the circle x = 0, y^{2} + z^{2} = 1.

**Solution**

We have x dx = y dy = z dz. Integrating, y^{2} = x^{2} + h, z^{2} = x^{2} + k (*). We are told that some point of the circle x = 0, y^{2} + z^{2} = 1 belongs to the curve, so h and k must be non-negative with sum 1. Hence the curve (*) lies in the surface 2 x^{2} + 1 = y^{2} + z^{2}, which is a one-sheet hyperboloid with the x-axis as an axis of symmetry. However, not all points of this surface lie on a curve (*). Clearly we require |y| >= |x| and |z| ≥ |x|. Equally, it is clear that this is a sufficient condition for we can then find h, k to satisfy (*).

© John Scholes

jscholes@kalva.demon.co.uk

5 Mar 2002