1. In the Cartesian coordinate system $x O y$, points $A$ and $B$ lie on the parabola $y^{2}=2 x$, satisfying $\overrightarrow{O A} \cdot \overrightarrow{O B}=-1, F$ is the focus of the parabola. Then the minimum value of $S_{\triangle O F A}+S_{\triangle O F B}$ is $\qquad$ .
$-1 . \frac{\sqrt{2}}{2}$. It is known that $F\left(\frac{1}{2}, 0\right)$. Let $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$. Then $x_{1}=\frac{y_{1}^{2}}{2}, x_{2}=\frac{y_{2}^{2}}{2}$.
Notice that,
$$
\begin{array}{l}
-1=\overrightarrow{O A} \cdot \overrightarrow{O B}=x_{1} x_{2}+y_{1} y_{2} \\
=\frac{1}{4}\left(y_{1} y_{2}\right)^{2}+y_{1} y_{2} .
\end{array}
$$

Thus, $y_{1} y_{2}=-2$.
Therefore, $S_{\triangle O F A}+S_{\triangle O F B}$
$$
\begin{array}{l}
=\frac{1}{2}|O F|\left|y_{1}\right|+\frac{1}{2}|O F|\left|y_{2}\right| \\
=\frac{1}{4}\left(\left|y_{1}\right|+\left|y_{2}\right|\right) \geqslant \frac{1}{2} \sqrt{\left|y_{1} y_{2}\right|}=\frac{\sqrt{2}}{2},
\end{array}
$$

where the equality holds if and only if $\left|y_{1}\right|=\left|y_{2}\right|=\sqrt{2}$, i.e., when point $A(1, \sqrt{2})$ and $B(1,-\sqrt{2})$ or $A(1,-\sqrt{2})$ and $B(1, \sqrt{2})$.
Thus, $S_{\triangle O F A}+S_{\triangle O F B}$ achieves the minimum value $\frac{\sqrt{2}}{2}$.