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

Answer: 2

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
$=\underset{0}{\overset{4a/m^2}{\int }}\left(\sqrt{4ax}-mx\right)dx$
Given $\underset{0}{\overset{4a/m^2}{\int }}\left(\sqrt{4ax}-mx\right)dx=\frac{a^2}{3}$
$\Rightarrow {\left[\frac{4\sqrt{a}}{3}.x^{\frac{3}{2}}-\frac{m}{2}{x}^2\right]}_0^{\frac{4a}{m^2}}=\frac{a^2}{3}$
$\Rightarrow \frac{8}{3}\frac{a^2}{m^3}=\frac{a^2}{3}$
$\Rightarrow m^3=8$
$\Rightarrow m=2$