2. In the Cartesian coordinate system $x O y$, points $A$ and $B$ lie on the parabola $y^{2}=4 x$, satisfying $\overrightarrow{O A} \cdot \overrightarrow{O B}=-4$, $F$ is the focus of the parabola. Then $S_{\triangle O F A} \cdot S_{\triangle O F B}=$ $\qquad$ .
Let $A\left(\frac{y_{1}^{2}}{4}, y_{1}\right), B\left(\frac{y_{2}^{2}}{4}, y_{2}\right)$, then $\overrightarrow{O A} \cdot \overrightarrow{O B}=\frac{y_{1}^{2} y_{2}^{2}}{16}+y_{1} y_{2}=-4$
$$
\Rightarrow y_{1}^{2} y_{2}^{2}+16 y_{1} y_{2}+64=0 \Rightarrow y_{1} y_{2}=-8 \text {. }
$$

Also, $F(1,0)$, so $S_{\triangle O F A} \cdot S_{\triangle O F B}=\frac{1}{2}\left|y_{1}\right| \cdot \frac{1}{2}\left|y_{2}\right|=\frac{1}{4}\left|y_{1} y_{2}\right|=2$.