Take axes so that $A = (-a, 0)$, $B = (a, 0)$ and $C = (k, b)$. Then the orthocenter lies on the line $x = k$. The line $AC$ has gradient $\frac{b}{k+a}$, so the perpendicular has gradient $-\frac{k+a}{b}$. Hence the altitude from $B$ has equation $y + \frac{(x-a)(k+a)}{b} = 0$. So the intersection is $x = k, y = -\frac{(k-a)(k+a)}{b}$. So the locus is all or part of the parabola $by = a^2 - x^2$. But we can get an orthocenter with any x-coordinate (by taking $C$ to have the same x-coordinate), so we can get all points on the parabola.