2. In the Cartesian coordinate system $x O y$, it is known that points $A$ and $B$ lie on the parabola $y^{2}=4 x$, and satisfy $\overrightarrow{O A} \cdot \overrightarrow{O B}=-4, F$ is the focus of the parabola. Then $S_{\triangle O F A} \cdot S_{\triangle O F B}=$ $\qquad$ .
2. 2 .

From the problem, we know the point $F(1,0)$.
Let $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$.
Thus, $x_{1}=\frac{y_{1}^{2}}{4}, x_{2}=\frac{y_{2}^{2}}{4}$.
Then $-4=\overrightarrow{O A} \cdot \overrightarrow{O B}=x_{1} x_{2}+y_{1} y_{2}$
$$
\begin{array}{l}
=\frac{1}{16}\left(y_{1} y_{2}\right)^{2}+y_{1} y_{2} \\
\Rightarrow \frac{1}{16}\left(y_{1} y_{2}+8\right)^{2}=0 \\
\Rightarrow y_{1} y_{2}=-8 .
\end{array}
$$

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