Question

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

• A. $2$
• B. $-1$
• C. $\frac{1}{2}$
• D. none of these

Answer 1

The correct option is A $2$

Given,

$y^2=4ax,y=mx$ curves intersect at $(4\frac{a}{m^2},\frac{a}{m})$

Area enclosed is,

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

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

${[2\sqrt{a}\frac{x^{\frac{3}{2}}}{\frac{3}{2}}-m\frac{x^2}{2}]}_0^{4\frac{a}{m^2}}=\frac{a^2}{3}$

$\frac{4\sqrt{a}}{3}(4\frac{a}{m^2}{)}^{\frac{3}{2}}-\frac{m}{2}{\left(4\frac{a}{m^2}\right)}^2-0-0=\frac{a^2}{3}$

$\frac{32a^2}{3m^3}-\frac{8a^2}{m^3}=\frac{a^2}{3}$

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

$m^3=8$

$\therefore m=2$