10. Given the function $y=\sqrt{2 x^{2}+2}$ whose graph is the curve $G$, the focus of curve $G$ is $F$. A line $I_{1}$ passing through $F$ intersects curve $G$ at points $A$ and $C$, and a line $l_{2}$ passing through $F$ intersects curve $G$ at points $B$ and $D$, with $\overrightarrow{A C} \cdot \overrightarrow{B D}=0$.
(1) Find the equation of curve $G$ and the coordinates of the focus $F$.
(2) Find the minimum value of the area $S$ of quadrilateral $A B C D$.
Solution: (1) $\frac{y^{2}}{2}-x^{2}=1(y \geq \sqrt{2}), F(0, \sqrt{3})$. $\qquad$ 5 points
(2) The slope of line $A C$ exists and is not 0, let it be $k$, then the slope of line $B D$ is $-\frac{1}{k}$.
$$
\left\{\begin{array}{l}
y^{2}-2 x^{2}=2, \\
y=k x+\sqrt{3}
\end{array} \Rightarrow\left(k^{2}-2\right) x^{2}+2 \sqrt{3} k x+1=0, \cdots \cdots 10\right. \text { points }
$$

Since the x-coordinates of points $A$ and $C$ are one positive and one negative, hence $\frac{1}{k^{2}-2}\frac{1}{2}\right)$, then
$$
S=\frac{1}{2}|A C| \cdot|B D|=\frac{4\left(1+k^{2}\right)^{2}}{\left(2-k^{2}\right)\left(2 k^{2}-1\right)} \geq \frac{4\left(1+k^{2}\right)^{2}}{\left[\frac{\left(2-k^{2}\right)+\left(2 k^{2}-1\right)}{2}\right]^{2}}=16\left(\frac{1}{2}<k^{2}<2\right) \cdots \cdots 20 \text { points }
$$