1st Putnam 1938 Problem B4 The parabola P has focus a distance m from the directrix. The chord AB is normal to P at A. What is the minimum length for AB? Solution
: 3√3 m. We may take the equation of P as 2my = x 2 . The gradient at the point A (a, a 2 /2m) is a/m, so the normal at (a, b) is (y - a 2 /2m) = -m/a (x - a). Substituting in 2my = x 2 , it meets P at (x, y) where x 2 + 2m 2 /a x - (2m 2 + a 2 ) = 0, so the other point B has x = -(2m 2 /a + a). Thus AB 2 = (2a + 2m 2 /a) 2 + 4m 2 (1 + m 2 /a 2 ) 2 = 4a 2 (1 + m 2 /a 2 ) 3 . Differentiating, we find the minimum is at a 2 = 2m 2 and is AB 2 = 27m 2 . 1st Putnam 1938 © John Scholes jscholes@kalva.demon.co.uk 5 Mar 2002