Let $A(-\lambda,\lambda^{2}),B(\lambda,\lambda^{2}),C(\mu,\mu^{2}),D(-\mu,\mu^{2})$. $ABCD$ is an isosceles trapezoid. Midpoint $M(0,1)$ of the circle, $MB^{2}=MC^{2} \Rightarrow \lambda_{2}+\mu^{2}=1$. Area $ABCD\ =\ (\lambda+\mu)(\lambda^{2}-\mu^{2})=(\lambda+\sqrt{(1-\lambda^{2}}\ )(2\lambda^{2}-1)=f(\lambda)$. Area has a maximimum if $f'(\lambda)=0$, if $\lambda=\frac{\sqrt{6}}{6}+\frac{\sqrt{3}}{3}$, then the area equals $\frac{4\sqrt{6}}{9}$.