Six, Given the hyperbola \( C_{1}: \frac{x^{2}}{a^{2}}-\frac{y^{2}}{2 a^{2}}=1(a>0) \), and the parabola \( C_{2} \) with 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.
(1) Given $F_{1}(-\sqrt{3} a, 0), \therefore C_{2}: y^{2}=-4 \sqrt{3} a x$.
From $\left\{\begin{array}{l}y^{2}=-4 \sqrt{3} a x, \\ 2 x^{2}-y^{2}=2 a^{2} \text {. }\end{array}\right.$ Eliminating $y$, we get $x^{2}+2 \sqrt{3} a x-a^{2}=0$.
Since $\Delta=(2 \sqrt{3} a)^{2}+4 a^{2}>0, \therefore$ the equation (1) has two real roots $x_{1}, x_{2}$.
Also, $x_{1} \cdot x_{2}=-a^{2}<0$, so one of $x_{1}, x_{2}$ is positive and the other is negative. When $x_{1}>0, x_{2}>0$, $y^{2}=-4 \sqrt{3} a x$ has no real roots;
When $x_{2}<0$, $y^{2}=-4 \sqrt{3} a x$ has two distinct real roots, thus $C_{1}$ and $C_{2}$ have two distinct intersection points.
(2) Assume a chord $A B$ satisfying the conditions exists.
( I ) When the slope $k$ of the line exists, it is easy to see that $k \neq 0$. Let the equation of line $A B$ be $y=k(x+\sqrt{3} a)$.
From the system of equations $\left\{\begin{array}{l}y=k(x+\sqrt{3} a), \\ y^{2}=-4 \sqrt{3} a x\end{array}\right.$ Eliminating $y$, we get
$$
\begin{array}{l}
k^{2} x^{2}+2 \sqrt{3} a\left(k^{2}+2\right) x+3 a^{2} k^{2}=0 . \\
\because\left\{\begin{array}{l}
x_{1}+x_{2}=\frac{-2 \sqrt{3} a\left(k^{2}+2\right)}{k^{2}} \\
x_{1} x_{2}=3 a^{2}
\end{array}\right.
\end{array}
$$
$$
\therefore A B=\sqrt{\left(1+k^{2}\right)\left[\frac{12 a^{2}\left(k^{2}+2\right)^{2}-12 a^{2} k^{4}}{k^{4}}\right]}=\frac{4 \sqrt{3} a\left(k^{2}+1\right)}{k^{2}} \text {. }
$$

Also, the distance from the origin to the line $A B$ is $h=\frac{\sqrt{3} a|k|}{\sqrt{k^{2}+1}}$,
$$
\therefore S_{\triangle A O B}=\frac{1}{2} \cdot \frac{\sqrt{3} a|k|}{\sqrt{k^{2}+1}} \cdot \frac{4 \sqrt{3} a\left(k^{2}+1\right)}{k^{2}}=6 a^{2} \sqrt{1+\frac{1}{k^{2}}} \text {. }
$$
(II) When the slope $k$ of the line does not exist, i.e., $A B \perp x$ axis, we have $S_{\triangle A O B}=\frac{1}{2} \cdot 4 \sqrt{3} a \cdot \sqrt{3} a=6 a^{2}$,
$$
\because 6 a^{2} \sqrt{1+\frac{1}{k^{2}}} \geqslant 6 a^{2} \quad(k \neq 0),
$$
$\therefore S_{\triangle A O B}$ has a minimum value of $6 a^{2}$, at which the equation of line $A B$ is $x=-\sqrt{3} a$. When $k \rightarrow 0$, $\frac{1}{k^{2}} \rightarrow \infty, \therefore S_{\triangle A O B}$ has no maximum value.