Conics Strike Again!

The parabola has equation $y^2 = 4ax$ with focus at $(a,0)$ and latus rectum meeting the parabola at $(a,\pm 2a)$ .

The point of intersection of the two chords is $(a,2a)$ , one of the two points where the parabola meets the latus rectum. If the other endpoint of one of these chords has coordinates $(x,2\sqrt{ax})$ , then the midpoint of the chord has coordinates $(\tfrac{x+a}{2},a + \sqrt{ax})$ . For there to be two such chords, we want the equation $\tfrac12(x+a) + a + \sqrt{ax} \; = \; 1$ to have at least two solutions. After some simple algebra, this equation becomes $x^2 + 2(a-2)x + (3a-2)^2 \; = \; 0$ or $(x + a - 2)^2 \; = \; 8a(1-a)$ and hence we require $0 < a < 1$ . Since the product of the two roots is $(3a-2)^2$ , it is clear that both of these roots are nonnegative (and so each defines a chord on the parabola) for any $0 < a < 1$ .

Since the length of the latus rectum is $4a$ , the possible integer values of $4a$ are $1,2,3$ .

The answer is $1+2+3=\boxed{6}$ .