Question

The area bounded by $y^2=4ax$ and $y=mx$ is $\frac{a^2}{3}$ sq. units then $m$

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

The correct option is B 2

The two curves $y^2=4ax$ and $y=mx$ intersect at $(\frac{4a}{m^2},\frac{4a}{m})$
and the area enclosed by the two curves are given by

${\int }_0^{\frac{4a}{m^2}}(\sqrt{4ax}-mx)dx$

$\therefore {\int }_0^{\frac{4a}{m^2}}(\sqrt{4ax}-mx)dx=\frac{a^2}{3}$

$\Rightarrow \frac{8}{3}.\frac{a^2}{m^3}=\frac{a^2}{3}\Rightarrow m^3=8\Rightarrow m=2$