# 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 :   Option 1) $2\sqrt6$ Option 2) 48 Option 3) 24 Option 4) $4\sqrt3$

the parabola $y^{2}=4\lambda x$ and  the line $y=\lambda x$

If $\lambda>0$, then

Area = $\int_{0}^{4/\lambda}(2\sqrt\lambda\sqrt x-\lambda x)dx=\frac{1}{9}$

=> $[\frac{2\sqrt \lambda x^{\frac{3}{2}}}{3/2}-\frac{\lambda x^{2}}{2}]^{4/\lambda}_{0}=\frac{1}{9}$

=> $\frac{4}{3}\sqrt\lambda \frac{8}{\lambda ^{\frac{3}{2}}}-\lambda \frac{8}{\lambda ^{2}}=\frac{1}{9}$

=> $\frac{32}{3\lambda }-\frac{8}{\lambda }=\frac{1}{9}$

=> $\frac{8}{3\lambda }=\frac{1}{9}$

=> $\lambda=24$

Option 1)

$2\sqrt6$

Option 2)

48

Option 3)

24

Option 4)

$4\sqrt3$

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