### 2nd Putnam 1939

Problem B4

The axis of a parabola is its axis of symmetry and its vertex is its point of intersection with its axis. Find: the equation of the parabola which touches y = 0 at (1,0) and x = 0 at (0,2); the equation of its axis; and its vertex.

Solution

Straightforward.

The general equation of the a parabola is:   (ax + by)2 + cx + dy + e = 0. Its intersection with y = 0 is given by a2x2 + cx + e = 0. This must have a double root, so c = -2a2, e = -a2. Considering the other tangent, we find: d = -4b2, e = 4b2. So (up to an irrelevant constant factor) we have: a = 2, b = 1, c = -8, d = -4, e = 4; or a = 2, b = -1, c = -8, d = -4, e = 4. But in the first case the equation can be written as (2x + y - 2)2 = 0, which is a double line. It is debatable whether this qualifies as a parabola, but it would not normally be said to touch the points (1,0) and (0,2). So we are left with the parabola: (2x - y)2 - 8x - 4y + 4 = 0.

We want to put this in the form u = kv2. The line x + 2y = 0 is perpendicular to the line 2x - y = 0, so we change variables to X = 2x - y, Y = x + 2y, giving: X2 - 16/5 Y - 12/5 X + 4 = 0, or 16/5 (Y - 4/5) = (X - 6/5)2, which is the equation of a parabola with vertex X = 6/5, Y = 4/5, axis X = 6/5. Changing back to the original coordinates, x = (2X + Y)/5, y = (2Y - X)/5, the vertex is (16/25, 2/25) and the axis is 10x - 5y = 6.