5th Putnam 1942

Problem A4

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?



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'/x2 - 2y/x3 = 3/x2. Integrating y = bx2 - 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 X2 = 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-11/2.



5th Putnam 1942

© John Scholes
5 Mar 2002