4. A line $l$ with a slope of 2 is drawn through the focus $F$ of the parabola $y^{2}=8 x$. If $l$ intersects the parabola at points $A$ and $B$, then the area of $\triangle O A B$ is $\qquad$
$4.4 \sqrt{5}$.
The focus of the parabola is $F(2,0), l_{A B}: y=2 x-4$.
Substituting into the parabola equation yields
$$
y^{2}-4 y-16=0 \text {. }
$$

Let $A\left(x_{1}, y_{1}\right) 、 B\left(x_{2}, y_{2}\right)$. Then,
$$
\begin{array}{l}
y_{1}+y_{2}=4, y_{1} y_{2}=-16 . \\
\text { Thus }\left(y_{1}-y_{2}\right)^{2}=\left(y_{1}+y_{2}\right)^{2}-4 y_{1} y_{2}=80 \\
\Rightarrow\left|y_{1}-y_{2}\right|=4 \sqrt{5} .
\end{array}
$$

Therefore, $S_{\triangle O A B}=S_{\triangle O A F}+S_{\triangle O B F}$
$$
=\frac{1}{2} A F\left|y_{1}-y_{2}\right|=4 \sqrt{5} .
$$