Five. (20 points) Given a quadratic function whose graph opens upwards and does not pass through the origin $O$, with the vertex coordinates at $(1,-2)$, and intersects the $x$-axis at points $A$ and $B$, and the $y$-axis at point $C$, and satisfies the relationship $\mid O C \mid^{2} = |O A| \cdot |O B|$.
(1) Find the analytical expression of the quadratic function;
(2) Find the area of $\triangle A B C$.
Let the analytical expression of the quadratic function be
$$
y=a(x-1)^{2}-2=a x^{2}-2 a x+a-2,
$$

and $a>0, a \neq 2$, with the graph as shown in Figure 7.

The intersections of the graph with the $x$-axis and $y$-axis are $A\left(x_{1}, 0\right)$, $B\left(x_{2}, 0\right)$, and $C(0, a-2)$.
(1) From $|O C|^{2}=|O A| \cdot|O B|$, we have
$$
\begin{array}{l}
(a-2)^{2}=\left|x_{1} x_{2}\right| \\
=\left|\frac{a-2}{a}\right|,
\end{array}
$$

which simplifies to $a^{3}-4 a^{2}+4 a=|a-2|$.
When $0 < a < 2$, we have $a^{3}-4 a^{2}+5 a-2=0$. Thus, $(a-1)\left(a^{2}-3 a+2\right)=0$.

This gives $a_{1}=1$, $a_{2}=2$ (discard), and $a_{3}=1$ (discard).
When $a > 2$, we have $a^{3}-4 a^{2}+3 a+2=0$. Thus, $(a-2)\left(a^{2}-2 a-1\right)=0$.

This gives $a_{1}=2$ (discard), $a_{2}=1+\sqrt{2}$, and $a_{3}=1-\sqrt{2}<0$ (discard).
Therefore, $a=1+\sqrt{2}$,
$$
y=(1+\sqrt{2}) x^{2}-(2+2 \sqrt{2}) x+\sqrt{2}-1.
$$

Thus, the required function expression is
$$
y=x^{2}-2 x-1
$$

or $y=(1+\sqrt{2}) x^{2}-(2+2 \sqrt{2}) x+\sqrt{2}-1$.
(2) From $S_{\triangle A B C}=\frac{1}{2}|A B| \cdot|O C|$, we have the following two cases:

When $y=x^{2}-2 x-1$,
$|A B|=\left|x_{1}-x_{2}\right|$
$=\sqrt{\left(x_{1}+x_{2}\right)^{2}-4 x_{1} x_{2}}=2 \sqrt{2}$.
Also, $|O C|=1$, so $S_{\triangle A B C}=\frac{1}{2} \times 2 \sqrt{2} \times 1=\sqrt{2}$.
When $y=(1+\sqrt{2}) x^{2}-(2+2 \sqrt{2}) x+\sqrt{2}-1$ (graph not shown),
$$
\begin{array}{l}
|A B|=\left|x_{1}-x_{2}\right| \\
=\sqrt{\left(x_{1}+x_{2}\right)^{2}-4 x_{1} x_{2}}=2 \sqrt{2(\sqrt{2}-1)} .
\end{array}
$$

Also, $|O C|=\sqrt{2}-1$, so
$$
\begin{array}{l}
S_{\triangle A B C}=\frac{1}{2} \times 2 \sqrt{2(\sqrt{2}-1)} \times(\sqrt{2}-1) \\
=(\sqrt{2}-1) \sqrt{2(\sqrt{2}-1)} .
\end{array}
$$

Therefore, the area of $\triangle A B C$ is $(\sqrt{2}-1) \sqrt{2(\sqrt{2}-1)}$ or $\sqrt{2}$.