If the area (in sq. units) bounded by the parabola $y^2=4\lambda x$ and the line $y=\lambda x,\lambda >0,$ is $\frac{1}{9}$, then $\lambda$ is equal to

The given equations are $y^2=4\lambda x$ and $y=\lambda x$

Using above equations, we can write

${\left(\lambda x\right)}^2=4\lambda x\Rightarrow {\lambda }^2{x}^2-4\lambda x=0$

$\Rightarrow \lambda x\left(\lambda x-4\right)=0$

$\Rightarrow x=0 \& x=\frac{4}{\lambda }$

Area $=\frac{1}{9}=\underset{0}{\overset{\frac{4}{\lambda }}{\int }}\left(\sqrt{4\lambda x}-\lambda x\right)dx$

$\Rightarrow {\left(\frac{\sqrt{4\lambda }{x}^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}-\frac{\lambda x^2}{2}\right)}_0^{\frac{4}{\lambda }}=\frac{1}{9}$

$\Rightarrow \lambda =24$

Hence, the required value is $24$.