Question

If area bounded by the curves $y^{2} = 4ax$ and $y = mx$ is $\frac{a^{2}}{3}$, then the value of $m$ is

Answer 1

Given curves $y^{2} = 4ax$ and $y = mx$ intersect at
$\left(\frac{4a}{m^{2}}, \frac{4a}{m}\right)$ The area enclosed by the two curves is
$=\int\limits_{0}^{4a/ m^2} \left(\sqrt{4ax}-mx\right)dx $
Given $\int\limits_{0}^{4a /m^2} \left(\sqrt{4ax}-mx\right)dx = \frac{a^{2}}{3}$
$\Rightarrow \left[\frac{4\sqrt{a}}{3}. x^{\frac{3}{2}}-\frac{m}{2} x^{2}\right]_{0}^{\frac{4a}{m^{2}}} = \frac{a^{2}}{3}$
$\Rightarrow \frac{8}{3} \frac{a^{2}}{m^{3}}= \frac{a^{2}}{3} $
$\Rightarrow m^{3} = 8 $
$\Rightarrow m=2$