For y^2 = 4ax, any point is P(t) = (at^2, 2at). This comes from setting y = 2at and computing x = y^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.