For $y^2$=4ax, a point (x1,y1) is inside the parabola if $y1^{2-4ax1}$ < 0, on it if $y1^{2-4ax1}$ = 0, and outside if $y1^{2-4ax1}$ > 0. Note: "inside" the parabola means the region containing the focus. To find the parameter t of a point on the parabola: from y = 2at, we get t = $\frac{y}{2a}$, then verify x = $at^2$.