13.4.31 * Given a line $l$ intersects the ellipse $C: \frac{x^{2}}{2}+y^{2}=1$ at points $A, B$, and $O$ is the origin.
(1) Find the maximum value of the area of $\triangle A O B$, and the equation of the line $l$ at this time;
(2) Let $L=\{l \mid$ line $l$ makes the area of $\triangle A O B$ reach its maximum value $\}, l_{1} 、 l_{2} 、 l_{3} 、 l_{4} \in L$ and satisfy $l_{1} / / l_{2}, l_{3} / / l_{4}, k_{1} 、 k_{2} 、 k_{3} 、 k_{4}$ are the slopes of the corresponding lines and $k_{1}+k_{2}+\dot{k}_{3}+k_{4}=0$, find the minimum value of the area of the quadrilateral formed by these four lines.
(1) When the slope of $l$ does not exist, we can set $l: x=t$, then
$$
S_{\triangle A O B}=\frac{1}{2}|t| \cdot 2 \sqrt{1-\frac{t^{2}}{2}}=\frac{\sqrt{2}}{2} \sqrt{t^{2}\left(2-t^{2}\right)} \leqslant \frac{\sqrt{2}}{2}
$$

Equality holds if and only if $t^{2}=2-t^{2}$, i.e., $t= \pm 1$.
When the slope of $l$ exists, we can set $l: y=k x+m$, from $\left\{\begin{array}{l}y=k x+m, \\ x^{2}+2 y^{2}=2,\end{array}\right.$ we get $\left(1+2 k^{2}\right) x^{2}+$ $4 k m x+2\left(m^{2}-1\right)=0$, so
$$
\Delta=16 k^{2} m^{2}-8\left(1+2 k^{2}\right)\left(m^{2}-1\right)=8\left(1+2 k^{2}-m^{2}\right),
$$

Therefore, $S_{\triangle A O B}=\frac{1}{2}|A B| \cdot h$. (where $h$ represents the distance from $O$ to $A B$)
$$
\begin{array}{l}
=\frac{1}{2} \sqrt{1+k^{2}} \cdot \frac{\sqrt{8\left(1+2 k^{2}-m^{2}\right)}}{1+2 k^{2}} \cdot \frac{|m|}{\sqrt{1+k^{2}}} \\
=\sqrt{2} \frac{\sqrt{\left(1+2 k^{2}-m^{2}\right) m^{2}}}{1+2 k^{2}} \leqslant \sqrt{2} \cdot \frac{\frac{1}{2}\left(1+2 k^{2}\right)}{1+2 k^{2}}=\frac{\sqrt{2}}{2},
\end{array}
$$

Equality holds if and only if $m^{2}=1+2 k^{2}-m^{2}$, i.e., $m^{2}=k^{2}+\frac{1}{2}$.
In summary, the maximum value of $S_{\triangle A O B}$ is $\frac{\sqrt{2}}{2}$, at this time the equation of the line $l$ is $x= \pm 1$ or $y=$ $k x \pm \sqrt{k^{2}+\frac{1}{2}}$.
(2) By symmetry, the lines
$$
\begin{array}{c}
l_{1}: y=k x+\sqrt{k^{2}+\frac{1}{2}},(k>0) l_{2}: y=k x-\sqrt{k^{2}+\frac{1}{2}}, \\
l_{3}: y=-k x+\sqrt{k^{2}+\frac{1}{2}}, l_{4}: y=-k x-\sqrt{k^{2}+\frac{1}{2}},
\end{array}
$$

form a rhombus, so $S=4 \cdot \frac{1}{2} \sqrt{k^{2}+\frac{1}{2}} \cdot \frac{\sqrt{k^{2}+\frac{1}{2}}}{k}=$ $2\left(k+\frac{1}{2 k}\right) \geqslant 2 \sqrt{2}$, equality holds if and only if $k=\frac{1}{2 k}$, i.e., $k=\frac{\sqrt{2}}{2}$, $S_{\min }=2 \sqrt{2}$.