14. Given: Hyperbola $C_{1}: \frac{x^{2}}{a^{2}}-\frac{y^{2}}{2 a^{2}}=1(a>0)$, parabola $C_{2}$ has its vertex at the origin $O$, and the foci of $C_{1}$ are the left focus $F_{1}$ of $C_{2}$. Does there exist a chord $A B$ passing through the focus $F_{1}$ of $C_{2}$ such that the area $S_{\triangle A O B}$ of $\triangle A O B$ has a maximum or minimum value? If so, find the equation of the line $A B$ and the extremum; if not, explain the reason.
14. Assuming a chord $AB$ that meets the conditions exists.
(i) When the slope $k$ of the line exists, it is easy to know that $k \neq 0$. Let the equation of line $AB$ 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 $y$ yields $k^{2} x^{2}+2 \sqrt{3} a\left(k^{2}+2\right) x+3 a^{2} k^{2}=0$. Since $\left\{\begin{array}{l}x_{1}+x_{2}=\frac{-2 \sqrt{3} a\left(k^{2}+2\right)}{k^{2}} \\ x_{1} x_{2}=3 a^{2}\end{array}\right.$, we have $|AB|=\sqrt{\left(1+k^{2}\right)\left[\frac{12 a^{2}\left(k^{2}+2\right)^{2}-12 a^{2} k^{4}}{k^{4}}\right]}=\frac{4 \sqrt{3} a\left(k^{2}+1\right)}{k^{2}}$. The distance from the origin to line $AB$ is $h=\frac{\sqrt{3} a|k|}{\sqrt{k^{2}+1}}$. Therefore, $S_{\triangle AOB}=\frac{1}{2} \cdot \frac{\sqrt{3} a|k|}{\sqrt{k^{2}+1}} \cdot \frac{4 \sqrt{3} a\left(k^{2}+1\right)}{k^{2}}=6 a^{2} \sqrt{1+\frac{1}{k^{2}}}$.
(ii) When the slope $k$ of the line does not exist, i.e., $AB \perp x$-axis, we have $S_{\triangle AOB}=\frac{1}{2} \cdot 4 \sqrt{3} a \cdot \sqrt{3} a=6 a^{2}$. Since $6 a^{2} \sqrt{1+\frac{1}{k^{2}}} \geqslant 6 a^{2} (k \neq 0)$, the minimum value of $S_{\triangle AOB}$ is $6 a^{2}$, at which point the equation of line $AB$ is $x=-\sqrt{3} a$. In summary, when $k \rightarrow 0$, $\frac{1}{k^{2}} \rightarrow \infty$, thus $S_{\triangle AOB}$ has no maximum value.