Six, (Full score 1.2 outside) The ellipse $C_{1}: \frac{x^{2}}{a^{2}} + \frac{y^{2}}{2 a^{2}}=1(a>0)$, the parabola $C_{2}$ has its vertex at the origin $O$, and the focus of $C_{2}$ is the left focus $F_{1}$ of $C_{1}$.
(1) Prove that $C_{1}$ and $C_{2}$ always have two different intersection points;
(2) Does there exist a chord $A B$ passing through the focus $F_{1}$ of $C_{2}$ such that the area of $\triangle A O B$ has a maximum or minimum value? If it exists, find the equation of the line $A B$ and the size of the maximum or minimum value; if not, explain the reason.
$$
=\frac{4 \sqrt{3} a\left(k^{2}+1\right)}{k^{2}} .
$$

The distance from the origin to the line $AB$ is $h=\frac{\sqrt{3} a|k|}{\sqrt{k^{2}+1}}$.
$$
\begin{aligned}
\therefore S_{\triangle A O B} & =\frac{1}{2} \cdot \frac{\sqrt{3} a|k|}{\sqrt{k^{2}+1}} \cdot \frac{4 \sqrt{3} a\left(k^{2}+1\right)}{k^{2}} \\
& =6 a^{2} \sqrt{1+\frac{1}{k^{2}}} .
\end{aligned}
$$
(ii) When the slope of the line $k$ does not exist, i.e., $AB \perp x$-axis, we have $S_{\text {Mom }}=\frac{1}{2} \cdot 4 \sqrt{3} a \cdot \sqrt{3} a=6 a^{2}$.
$$
\because 6 a^{2} \sqrt{1+\frac{1}{k^{2}}} \geqslant 6 a^{2},(k \neq 0)
$$
$\therefore$ The minimum value of $S_{\triangle A C B}$ is $6 a^{2}$, at which time the equation of the line $AB$ is $x=-\sqrt{3} a$.
When $k \rightarrow 0$, $\frac{1}{k^{2}} \rightarrow \infty$, therefore, $S_{\triangle M A}$ reaches its maximum value.