5th Putnam 1942 Problem B2 Let P a be the parabola y = a 3 x 2 /3 + a 2 x/2 - 2a. Find the locus of the vertices of P a , and the envelope of P a . Sketch the envelope and two P a .
We can write the equation of P a as (y + 35a/16) = (a 3 /3)(x + 3/4a) 2 , so the vertex is x = -3/4a, y = -35a/16. The locus of the vertex is xy = 105/64. The graph shows P 3 , P 2 , P 1 , P 1/2 , P 1/3 and the two hyperbolae yx = -7/6, yx = 10/3. It shows that for positive a, the parabolas touch the lower branches of the hyperbolae. For negative a they touch the upper branches. That is not hard to verify. We claim that P a and xy = 10/3 touch at x = -2/a, y = -5a/3. The point obviously lies on xy = 10/3. We have (a 3 /3)(x + 3/4a) 2 = (1/3)(-2 + 3/4) 2 a = 25a/48 = (-5a/3 + 35a/16), so the point also lies on P a . The gradient of xy = 10/3 at the point is -10/(3x 2 ) = -5a 2 /6. The gradient of P a at the point is 2a 3 x/3 + a 2 /2 = -5a 2 /6. Similarly, we claim that P a and xy = -7/6 touch at x = 1/a, y = -7a/6. The point obviously lies on xy = -7/6. We have (a 3 /3)(x + 3/4a) 2 = (a/3)(1 + 3/4) 2 = 49a/48 = (-7a/6 + 35a/16), so the point also lies on P a . The gradient of xy = -7/6 at the point is 7/(6x 2 ) = 7a 2 /6. The gradient of P a at the point is 2a 3 x/3 + a 2 /2 = a 2 (2/3 + 1/2) = 7a 2 /6. It is less clear how you get the hyperbolas. One standard approach is to look for the singular points of the mapping f(a,t) = (t, a 3 t 2 /3 + a 2 t/2 - 2a). The matrix for the derivative is: 1 0 2a 3 t/3+a 2 /2 a 2 t 2 +at-2 which has zero determinant when at = 1 or -2, so xy = 7/6 or -10/3. Thanks to Dave Rusin 5th Putnam 1942 © John Scholes jscholes@kalva.demon.co.uk 20 January 2004 Last corrected/updated 20 Jan 04