SAT1000 - P787

The parabola $y^2 = 2x$ has a focus of $F(\frac{1}{2}, 0)$ , so $F_x = -\frac{1}{2}$ and $F_y = 0$ , and a directrix of $x = -\frac{1}{2}$ , so $P_x = Q_x = -\frac{1}{2}$ .

Let $p = P_y = A_y$ and $q = -Q_y = -B_y$ . Since $A$ and $B$ are on the parabola $y^2 = 2x$ , $A_x = \frac{1}{2}p^2$ and $B_x = \frac{1}{2}q^2$ .

The area of $\triangle PQF$ is $A_{\triangle PQF} = \frac{1}{2}(p + q)$ . The area of $\triangle ABF$ is the area of trapezoid $PQBA$ minus the sum of $A_{\triangle PAF}$ , $A_{\triangle PQF}$ , and $A_{\triangle BQF}$ ; or $A_{\triangle ABF} = \frac{1}{2}(\frac{1}{2}(p^2 + 1) + \frac{1}{2}(q^2 + 1))(p + q) - (\frac{1}{2}(p + q) + \frac{1}{2}(\frac{1}{2}(p^2 + 1))p) + \frac{1}{2}(\frac{1}{2}(q^2 + 1))q)$ . Equating $A_{\triangle PQF} = 2A_{\triangle ABF}$ and simplifying solves to $q = \frac{2}{p}$ .

Substituting $q = \frac{2}{p}$ into $B_x$ and $B_y$ gives $B(\frac{2}{p^2}, -\frac{2}{p})$ .

The midpoint $M$ of $A$ and $B$ is then $M = (\frac{1}{2}(A_x + B_x), \frac{1}{2}(A_y + B_y)) = (\frac{p^4 + 4}{4p^2}, \frac{p^2 - 2}{2p})$ , so the locus of point $M$ is $x = \frac{p^4 + 4}{4p^2}$ and $y = \frac{p^2 - 2}{2p}$ . Rearranging these two equations to eliminate $p$ gives $f(x, y) = y^2 - x + 1 = 0$ .

When $y = 50$ , the only possible value for $x$ is $x = y^2 + 1 = 50^2 + 1 = \boxed{2501}$ .