10. A line passing through the origin with a positive slope intersects the ellipse $\frac{x^{2}}{4}+y^{2}=1$ at points $E, F$. Let $A(2,0), B(0,1)$. Find the maximum value of the area of quadrilateral $A E B F$.
As shown in the figure, let $E(2 \cos \theta, \sin \theta), F(-2 \cos \theta, -\sin \theta)$, where $\theta \in \left(0, \frac{\pi}{2}\right)$. Then $S_{A E B F} = S_{\triangle A E F} + S_{\triangle B E F} = 2 \sin \theta + 2 \cos \theta = 2 \sqrt{2} \sin \left(\theta + \frac{\pi}{4}\right)$. Therefore, when $\theta = \frac{\pi}{4}$, that is, when the slope of line $E F$ is 1, the maximum area of quadrilateral $A E B F$ is $2 \sqrt{2}$.