10. (20 points) Given $P\left(2 a^{2}, 4 a\right)$ (where $a$ is a non-zero constant) is a point on the parabola $y^{2}=8 x$. Two lines with complementary inclination angles are drawn through $P$ and intersect the parabola at two other points $A$ and $B$.
(1) Prove that the slope of line $A B$ is a constant;
(2) If point $P$ is above the $x$-axis, and $A$ and $B$ are below the $x$-axis, find the minimum upper bound of the area of $\triangle P A B$.
10. (1) Let points $A\left(2 t_{1}^{2}, 4 t_{1}\right)$ and $B\left(2 t_{2}^{2}, 4 t_{2}\right)$.

Given that the slopes of lines $PA$ and $PB$ are complementary,
$$
\begin{array}{l}
\Rightarrow k_{P A}+k_{P B}=0 \\
\Rightarrow \frac{4 t_{1}-4 a}{2 t_{1}^{2}-2 a^{2}}+\frac{4 t_{2}-4 a}{2 t_{2}^{2}-2 a^{2}}=0 \\
\Rightarrow \frac{2}{t_{1}+a}+\frac{2}{t_{2}+a}=0 \\
\Rightarrow t_{1}+t_{2}=-2 a \\
\Rightarrow k_{A B}=\frac{4 t_{2}-4 t_{1}}{2 t_{2}^{2}-2 t_{1}^{2}}=\frac{2}{t_{1}+t_{2}}=-\frac{1}{a} .
\end{array}
$$

Thus, the slope of line $AB$ is a constant $-\frac{1}{a}$.
(2) According to the problem, $a>0$, and $t_{1}, t_{2}<0$.

From (1), we have $t_{1}+t_{2}=-2 a$.
Therefore, $t_{1}, t_{2} \in(-2 a, 0)$.
Given points $P\left(2 a^{2}, 4 a\right)$, $A\left(2 t_{1}^{2}, 4 t_{1}\right)$, and $B\left(2 t_{2}^{2}, 4 t_{2}\right)$, we get
$$
\begin{array}{l}
S_{\triangle P A B}=\left|\frac{1}{2}\right| \begin{array}{ccc}
2 a^{2} & 4 a & 1 \\
2 t_{1}^{2} & 4 t_{1} & 1 \\
2 t_{2}^{2} & 4 t_{2} & 1
\end{array}|| \\
=|4| \begin{array}{ccc}
a^{2} & a & 1 \\
\left(t_{1}^{2}-a^{2}\right) & \left(t_{1}-a\right) & 0 \\
\left(t_{2}^{2}-a^{2}\right) & \left(t_{2}-a\right) & 0
\end{array}|| \\
=\left|4\left(t_{1}-a\right)\left(t_{2}-a\right)\left(t_{1}-t_{2}\right)\right| \\
=8\left|\left(t_{1}-a\right)\left(t_{1}+a\right)\left(t_{1}+3 a\right)\right| .
\end{array}
$$

Consider the function
$$
f(t)=(t-a)(t+a)(t+3 a)(t \in(-2 a, 0)) \text {. }
$$

Since $f^{\prime}(x)=3 t^{2}+6 a t-a^{2}=3(t+a)^{2}-4 a^{2}<0$, we know that $f(t)$ is monotonically decreasing on $(-2 a, 0)$, i.e.,
$$
-3 a^{3}<f(t)<3 a^{3} \text {. }
$$

Thus, $S_{\triangle P A B}=8\left|\left(t_{1}-a\right)\left(t_{1}+a\right)\left(t_{1}+3 a\right)\right|<24 a^{3}$.
When $t_{1} \rightarrow 0$ or $-2 a$, $S_{\triangle P A B} \rightarrow 24 a^{3}$.
Therefore, the minimum upper bound of the area of $\triangle P A B$ is $24 a^{3}$.