If the area enclosed by $${y^2} = 4ax$$   and line $$y = ax$$  is $$\frac{1}{3}$$ sq. units , then the area enclosed by $$y = 4x$$  with same parabola is :

Solution :
Point of intersection of $${y^2} = 4ax$$   and $$y = ax$$  are $$\left( {0,\,0} \right)$$  and $$\left( {\frac{4}{a},\,4} \right)$$

$$\eqalign{ & {\text{Given,}}\,\int\limits_0^4 {\left[ {\frac{y}{a} - \frac{{{y^2}}}{{4a}}} \right]} dy = \frac{1}{3} \cr & \Rightarrow \frac{8}{a} - \frac{1}{{12a}} \times 64 = \frac{1}{3} \cr & \Rightarrow \frac{8}{{3a}} = \frac{1}{3} \cr & \Rightarrow a = 8 \cr} $$
So, the parabola is $${y^2} = 32x$$

Area enclosed by $$y = 4x$$   is $$\int\limits_0^8 {\left[ {\frac{y}{4} - \frac{{{y^2}}}{{32}}} \right]} dy = \left[ {\frac{{{y^2}}}{8} - \frac{{{y^3}}}{{96}}} \right]_0^8 = \frac{8}{3}$$