For $y^2$ = 4ax and the line y = mx (m > 0), intersection points are at (0, 0) and (4a/$m^2$, 4a/m). Using horizontal strips from y = 0 to y = 4a/m: the parabola gives x = y^$\frac{2}{4a}$ and the line gives x = $\frac{y}{m}$. Since y/m > y^$\frac{2}{4a}$ on (0, 4a/m), the area = integral from 0 to 4a/m of [y/m - y^$\frac{2}{4a}$] dy = [y^$\frac{2}{2m}$ - y^$\frac{3}{12a}$] from 0 to 4a/m = 16a^$\frac{2}{2m^3}$ - 64a^$\frac{3}{12a*m^3}$ = 8$a^2$/$m^3$ - 16a^$\frac{2}{3m^3}$ = 8a^$\frac{2}{3m^3}$. This standard result 8a^$\frac{2}{3m^3}$ appears frequently in JEE. For m = 1 and a = 1, the area is 8/3.