Let C be the family of conics (2y + x)^{2} = a(y + x). Find C', the family of conics which are orthogonal to C. At what angle do the curves of the two families meet at the origin?

**Solution**

For most points P in the plane we can find a unique conic in the family passing through the point. Thus we should be able to find the gradient of members of the family at (x, y) in a formula which is independent of a. We then use this to get a formula for the gradient of the orthogonal family and solve the resulting first-order differential equation to get the orthogonal family.

Thus we have 8y y' + 4x y' + 4y + 2x = ay' + a = (y' + 1)(2y + x)^{2}/(y + x). So y'(2y+x)( 4(x+y) - (2y+x) ) = (2y+x)^{2} - (x+2y)(x+y), or y'(2y+x)(2y+3x) = -2x(2y+x), so y' = -x/(2y+3x). Hence the orthogonal family satisfies y' = (2y+3x)/x. So y'/x^{2} - 2y/x^{3} = 3/x^{2}. Integrating y = bx^{2} - 3x. These are all parabolas.

All members of both families pass through the origin. Changing coordinates to X = x + 2y, Y = y - 2x, the equation of a member of the first family becomes X^{2} = a(3X-Y)/5 or Y = - 5/a (X - 3a/10)^{2} + 9a/20. This has gradient 3 (in the new system) at the origin. In the old system the tangent is y = -x. The orthogonal set obviously has gradient -3 at the origin. If the angle between them is k, then tan k = (-1 +3)/(1+3) = 1/2. So k = tan^{-1}1/2.

© John Scholes

jscholes@kalva.demon.co.uk

5 Mar 2002