### 11th Putnam 1951

Problem A6

Let C be a parabola. Take points P, Q on C such that (1) PQ is perpendicular to the tangent at P, (2) the area enclosed by the parabola and PQ is as small as possible. What is the position of the chord PQ?

Solution

We may take the parabola as y = x2 and P as (a, a2). The tangent at P has slope 2a, so the normal has slope -1/2a and equation (y - a2) + 1/2a (x - a) = 0. This meets the parabola at x2 + x/2a - a2 - 1/2 = 0, so the other root is x = -1/2a - a and Q is (-1/2a - a, (a + 1/2a)2). The area under the curve between P and Q is 1/3 (a3 + (a + 1/2a)3 ). The area under the chord PQ is (2a + 1/2a) (a2 + (a + 1/2a)2)/2 = 2a3 + 3a/2 + 1/2a + 1/16a3. Hence the area enclosed is 4/3 a3 + a + 1/4a + 1/48a3. We wish to minimise this. The gradient is 4a2 + 1 -1/4a2 - 1/16a4 = 1/16a4 (4a2 + 1)2(2a - 1)(2a + 1). Hence there is a minimum at a = 1/2 (and another at a = -1/2). In the minimum position P is (1/2, 1/4) or (-1/2, 1/4), so the minimum positions are the intersection of the parabola with the perpendicular to the axis through the focus (0, 1/4).