If the area (in sq. units) bounded by the parabola y ^ 2 = 4 and...

The given equations are ${y}^{2} = 4\lambdax$ and $y = \lambdax$

Using above equations, we can write

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

$⇒ \lambdax\left(\lambdax - 4\right) = 0$

$⇒ x = 0 \text{ and } x = \dfrac{4}{\lambda}$

Area $= \dfrac{1}{9} = \mathop{∫}_{0}^{\dfrac{4}{\lambda}}\left(\sqrt{4\lambdax} - \lambdax\right)dx$

$⇒ {\left(\dfrac{\sqrt{4\lambda}{x}^{\dfrac{3}{2}}}{\dfrac{3}{2}} - \dfrac{\lambda{x}^{2}}{2}\right)}_{0}^{\dfrac{4}{\lambda}} = \dfrac{1}{9}$

$⇒ \lambda = 24$

Hence, the required value is $24$.