9. (20 points) Let the parabola $y^{2}=2 p x(p>0)$ intersect the line $x+y=1$ at points $A$ and $B$. If $O A \perp O B$, find the equation of the parabola and the area of $\triangle O A B$.
Let points $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$.
From $y^{2}=2 p x$ and $x+y=1$, we get
$$
y^{2}+2 p y-2 p=0 \text {. }
$$

Thus, $\dot{x}_{1}=1+p-\sqrt{p^{2}+2 p}$,
$$
\begin{array}{l}
y_{1}=-p+\sqrt{p^{2}+2 p}, \\
x_{2}=1+p+\sqrt{p^{2}+2 p} \\
y_{2}=-p-\sqrt{p^{2}+2 p} .
\end{array}
$$

Since $O A \perp O B$, i.e., $\overrightarrow{O A} \cdot \overrightarrow{O B}=0$, we have
$$
\begin{array}{l}
x_{1} x_{2}+y_{1} y_{2}=0 \\
\Rightarrow\left[(1+p)^{2}-\left(p^{2}+2 p\right)\right]+\left[p^{2}-\left(p^{2}+2 p\right)\right]= \\
\Rightarrow 1-2 p=0 .
\end{array}
$$

Therefore, the equation of the parabola is $y^{2}=x$.
Thus, $A\left(\frac{3-\sqrt{5}}{2}, \frac{-1+\sqrt{5}}{2}\right), B\left(\frac{3+\sqrt{5}}{2}, \frac{-1-\sqrt{5}}{2}\right)$.
Then $O A^{2}=x_{1}^{2}+y_{1}^{2}=5-2 \sqrt{5}$,
$$
O B^{2}=x_{2}^{2}+y_{2}^{2}=5+2 \sqrt{5} \text {. }
$$

Therefore, $S_{\triangle O A B}=\frac{1}{2}|O A||O B|=\frac{\sqrt{5}}{2}$.