3. Given the parabola $y^{2}=2 p x(p>0)$, with its focus at $F$, a line passing through $F$ with an inclination angle of $\theta$ intersects the parabola at points $A$ and $B$. The line connecting point $A$ and the vertex $O$ of the parabola intersects the directrix at point $B^{\prime}$, and the line connecting point $B$ and $O$ intersects the directrix at point $A^{\prime}$. Find the area of quadrilateral $A B B^{\prime} A^{\prime}$.
3. First consider the case when $\theta=\frac{\pi}{2}$, at this point the required area is $2 p^{2}$ (obtained from the properties of the parabola).

When $\theta \neq \frac{\pi}{2}$, let the slope of the line be $k=\operatorname{tg} \theta, A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$, then $y_{1}^{2}=2 p x_{1}$, $y_{2}^{2}=2 p x_{2}$, and $\quad y_{1} y_{2}=-p, y_{1}+y_{2}=\frac{2 p}{k}$.

It is not difficult to find $B^{\prime}\left(-\frac{p}{2},-\frac{p^{2}}{y_{1}}\right)$, and since $B B^{\prime}$ is parallel to the $x$-axis, $A A^{\prime}$ is parallel to the $x$-axis, then $A A^{\prime} B^{\prime} B$ forms a right trapezoid. Therefore,
$$
\begin{aligned}
S_{A B B^{\prime} A^{\prime}} & =\frac{1}{2}\left|A^{\prime} B^{\prime}\right| \cdot|A B|=\frac{1}{2}\left|y_{1}-y_{2}\right| \sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}} \\
& =2 p^{2}\left(1+\frac{1}{k^{2}}\right)^{\frac{3}{2}} .
\end{aligned}
$$