10. (20 points) As shown in Figure 2, the right triangle $\triangle ABC$ is inscribed in the parabola $y^{2}=2 p x$, with the hypotenuse $BC \perp x$-axis at point $M$. Extend $MA$ to point $D$ such that the circle $\odot N$ with diameter $AD$ is tangent to the $x$-axis at point $E$. Connect $BE$, intersecting the parabola at point $F$. If the area of quadrilateral $AFBC$ is $8 p^{2}$, find the area of $\triangle ACD$.
10. It is known that $N E / / C M$.

Then $\angle E N A=\angle A M C$
$\Rightarrow$ isosceles $\triangle A N E \backsim$ isosceles $\triangle A M C$
$\Rightarrow \angle N A E=\angle C A M$
$\Rightarrow C 、 A 、 E$ are collinear.
By symmetry, quadrilateral $C A F B$ is an isosceles trapezoid.
Let $A\left(x_{1}, y_{1}\right), C\left(x_{2}, y_{2}\right)$.
Since $A C \perp A B$
$$
\begin{array}{l}
\Rightarrow \frac{y_{2}-y_{1}}{x_{2}-x_{1}} \cdot \frac{-y_{2}-y_{1}}{x_{2}-x_{1}}=-1 \\
\Rightarrow y_{2}^{2}-y_{1}^{2}=\left(x_{2}-x_{1}\right)^{2}=2 p\left(x_{2}-x_{1}\right) \\
\Rightarrow x_{2}-x_{1}=2 p \\
\Rightarrow y_{2}^{2}-y_{1}^{2}=4 p^{2} . \\
\text { Also } S_{\text {figure } F B C}=\frac{1}{2}\left(2 y_{1}+2 y_{2}\right)\left(x_{2}-x_{1}\right)=8 p^{2} \\
y_{1}+y_{2}=4 p .
\end{array}
$$

Also $S_{\text {figure } A F B C}=\frac{1}{2}\left(2 y_{1}+2 y_{2}\right)\left(x_{2}-x_{1}\right)=8 p^{2}$, hence

From equations (1) and (2), we get $y_{1}=\frac{3 p}{2}, y_{2}=\frac{5 p}{2}$.
By Rt $\triangle A D E \backsim \mathrm{Rt} \triangle A B C$
$$
\Rightarrow \frac{S_{\triangle A C D}}{S_{\triangle A C B}}=\frac{D E}{A B}=\frac{A E}{A C}=\frac{y_{1}}{y_{2}-y_{1}}=\frac{3}{2} \text {. }
$$

Also $S_{\triangle A C B}=\frac{1}{2} \times 2 y_{2}\left(x_{2}-x_{1}\right)=5 p^{2}$, therefore,
$$
S_{\triangle A C D}=\frac{15}{2} p^{2} .
$$