# Find the area enclosed between the parabola $$y^2 = 4ax$$ and the line $$y = mx.$$

Suppose we are given two curves represented by y=f(x);y=g(x) where $f(x)\geq g(x)$ in [a,b]
Here the points of interaction of these curves are given by x=a and x=b obtained by taking the common values of y from the equation of the two curves.
Hence dA=[f(x)-g(x)]dx and the given area A can be taken as
$A=\int_a^b[f(x)-g(x)]dx.$
Given $y^2=4ax$ and y=mx are the two functions
To find the points of intersection,let us solve the two equation
since $y^2=4ax$ and $y^2=m^2x^2$
$4ax=m^2x^2$
$\;4a=m^2x\Rightarrow x=\frac{4a}{m^2}$
Also $x=0\Rightarrow y=0$ and $x=\frac{4a}{m^2}\Rightarrow y=\frac{4a}{m}$.
so the points of intersection are (0,0) and $\bigg(\frac{4a}{m^2},\frac{4a}{m}\bigg)$
Now the required area A is given by
$A=\int_0^{\frac{4a}{m^2}}(y_2-y_1)dx.$
$A=\int_0^{\frac{4a}{m^2}}(2\sqrt {ax}-mx)dx.$
$\Rightarrow A=\begin{bmatrix}2\sqrt a\frac{x^{\frac{3}{2}}}{\frac{3}{2}}-\frac{mx^2}{2}\end{bmatrix}_0^{\frac{4a}{m^2}}$
Applying the limits we get,
$A=\frac{4}{3}\sqrt a\bigg(\frac{4a}{m^2}\bigg)^{\frac{3}{2}}-\frac{m}{2}\bigg(\frac{4a}{m^2}\bigg)^2$
$A=\frac{8\times4\sqrt a\times a\sqrt a}{3m^3}-\frac{m\times16\times a^2}{2\times m^4}$
$A=\frac{32a^2}{3m^3}-\frac{16a^2}{2m^3}=\frac{8a^2}{3m^3}sq.units$
Hence the required area is $\frac{8a^2}{3m^3}$sq.units.