I OBM

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.