Official solution
y2 = 4\[\lambda \]x and y = \[\lambda x\]
If \[\lambda \] > 0 then
Hence \[\int\limits_0^{4/\lambda } {(2\sqrt \lambda \sqrt x } - \lambda x)dx = {1 \over 9}\]
\[ \Rightarrow \] \[{\left( {{{2\sqrt \lambda {x^{3/2}}} \over {3/2}} - {{\lambda {x^2}} \over 2}} \right)^{{4 \over \lambda }}} = {1 \over 9}\]
\[ \Rightarrow \] \[{4 \over 3}\sqrt \lambda {8 \over {{\lambda ^{3/2}}}} - \lambda {8 \over {{\lambda ^2}}} = {1 \over 9}\]
\[ \Rightarrow \] \[{{32} \over {3\lambda }} - {8 \over \lambda } = {1 \over 9}\]
\[ \Rightarrow \] \[ {8 \over 3\lambda } = {1 \over 9}\]
\[ \Rightarrow \] \[\lambda \] = 24