$k y^2=2(y-x)$ $$ 2 y^2=k x $$ Point of intersection → $$ \begin{aligned} & k y^2=\left(y-\frac{2 y^2}{k}\right) \\ & y=0 k y=2\left(\frac{1-2 y}{k}\right) \\ & k y+\frac{4 y}{k}=2 \\ & y=\frac{2}{k+\frac{4}{k}}=\frac{2 k}{k^2+4} \\ & \frac{2 k}{k^2+4}\left(\left(y-\frac{k y}{2}\right)-\left(\frac{2 y^2}{k}\right)\right) \cdot d y \\ & =\frac{y^2}{2}-\left.\left(\frac{k}{2}+\frac{2}{k}\right) \cdot \frac{y^3}{3}\right|_0 ^{\frac{2 k}{k^2+4}} \\ & =\left(\frac{2 k}{k^2+4}\right)^2\left[\frac{1}{2}-\frac{k^2+4}{2 k} \times \frac{1}{3} \times \frac{2 k}{k^2+4}\right] \\ & =\frac{1}{6} \times 4 \times\left(\frac{1}{k+\frac{4}{k}}\right)^2 \end{aligned} $$ $$ \begin{aligned} & A \cdot M \geq G \cdot M \frac{\left(k+\frac{4}{k}\right)}{2} \geq 2 \\ & \mathrm{k}+\frac{4}{\mathrm{k}} \geq 4 \end{aligned} $$ Area is maximum when $\mathrm{k}=\frac{4}{\mathrm{k}}$ $$ \mathrm{k}=2,-2 $$