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^2/(4a) and the line gives x = y/m. Since y/m > y^2/(4a) on (0, 4a/m), the area = integral from 0 to 4a/m of [y/m - y^2/(4a)] dy = [y^2/(2m) - y^3/(12a)] from 0 to 4a/m = 16a^2/(2m^3) - 64a^3/(12a*m^3) = 8a^2/m^3 - 16a^2/(3m^3) = 8a^2/(3m^3). This standard result 8a^2/(3m^3) appears frequently in JEE. For m = 1 and a = 1, the area is 8/3.