13.6.17 ** Given the parabola $y^{2}=2 p x(p>0)$, with its focus at $F$, a line passing through the focus $F$ with an inclination angle of $\theta(0<\theta<\pi)$ intersects the parabola at points $A$ and $B$. Connecting $A O$ (where $O$ is the origin) intersects the directrix at point $B^{\prime}$. Connecting $B O$ intersects the directrix at point $A^{\prime}$. Find the area of quadrilateral $A B B^{\prime} A^{\prime}$.
When $\theta=\frac{\pi}{2}$, $S_{\text {quadrilateral } A B B^{\prime} A^{\prime}}=2 p^{2}$. When $\theta \neq \frac{\pi}{2}$, let $k=\tan \theta$, and suppose $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$, then from $y=k\left(x-\frac{p}{2}\right)$,
(1) $y^{2}=2 p x$,
(2) Eliminating $x$ gives, $y^{2}-\frac{2 p}{k} y-p^{2}=0$, so $y_{1}+y_{2}=\frac{2 p}{k}$,
$$
y_{1} y_{2}=-p^{2} \text {. }
$$

The equation of the line $A O$ is: $y=\frac{y_{1}}{x_{1}} x$, which is $y=\frac{2 p}{y_{1}} x$, so the coordinates of the intersection point of $A O$ with the directrix are $B^{\prime}\left(-\frac{p}{2},-\frac{p^{2}}{y_{1}}\right)$. From (3), we know that $y_{2}=-\frac{p^{2}}{y_{1}}$, so the y-coordinates of $B$ and $B^{\prime}$ are equal, thus $B B^{\prime} \perp x$-axis. Similarly, $A A^{\prime} \perp x$-axis, hence the quadrilateral $A B B^{\prime} A^{\prime}$ is a right trapezoid. Therefore, its area is
$$
\begin{aligned}
S_{\text {quadrilateral } A B B^{\prime} A^{\prime}} & =\frac{1}{2}\left(\left|A A^{\prime}\right|+\left|B B^{\prime}\right|\right) \cdot\left|A^{\prime} B^{\prime}\right|=\frac{1}{2}|A B| \cdot\left|A^{\prime} B^{\prime}\right| \\
& =\frac{1}{2} \sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}} \cdot\left|y_{2}-y_{1}\right| \\
& =\frac{1}{2}\left(y_{2}-y_{1}\right)^{2} \sqrt{1+\frac{1}{k^{2}}}=\frac{1}{2} \sqrt{1+\frac{1}{k^{2}}}\left[\left(y_{1}+y_{2}\right)^{2}-4 y_{1} y_{2}\right] \\
& =2 p^{2}\left(1+\frac{1}{k^{2}}\right)^{\frac{3}{2}}=2 p^{2}\left(1+\cot ^{2} \theta\right)^{\frac{3}{2}} .
\end{aligned}
$$