For $y^2$ = 4ax, any point is P(t) = ($at^2$, 2at). This comes from setting y = 2at and computing x = y^$\frac{2}{4a}$ = $at^2$. The parameter t can be any real number: t > 0 gives points in the upper half, t < 0 in the lower half, t = 0 is the vertex. For $x^2$ = 4ay, parametric form is (2at, $at^2$). The parametric form is preferred in JEE because it reduces two-variable problems to single-variable ones.