8. Given points $A(1,-1), B(4,0), C(2,2)$, the plane region $D$ consists of all points $P(x, y)$ that satisfy
$$
\overrightarrow{A P}=\lambda \overrightarrow{A B}+\mu \overrightarrow{A C}(1<\lambda \leqslant a, 1<\mu \leqslant b)
$$

If the area of region $D$ is 8, then the minimum value of $a+b$ is . $\qquad$
8. 4 .

As shown in Figure 3, extend $A B$ to point $N$, and extend $A C$ to point $M$, such that
$$
|A N|=a|A B|,|A M|=b|A C| .
$$

Construct $\square A B E C$ and $\square A N G M$. Then quadrilateral $E H G F$ is a parallelogram.

From the conditions, the region $D$ composed of points $P(x, y)$ is the shaded area in the figure, i.e., $\square E H G F$ (excluding the boundaries $E H$ and $E F$).
Notice that,
$$
\overrightarrow{A B}=(3,1), \overrightarrow{A C}=(1,3), \overrightarrow{B C}=(-2,2) \text {. }
$$

Thus, $|A B|=|A C|=\sqrt{10},|B C|=2 \sqrt{2}$.
Then $\cos \angle C A B=\frac{10+10-8}{2 \times \sqrt{10} \times \sqrt{10}}=\frac{3}{5}$
$$
\Rightarrow \sin \angle C A B=\frac{4}{5} \text {. }
$$

Therefore, $S_{\text {DEHGF }}$
$$
\begin{aligned}
= & (a-1) \sqrt{10}(b-1) \sqrt{10} \times \frac{4}{5}=8 \\
\Rightarrow & (a-1)(b-1)=1 \\
\Rightarrow & a+b=a+\left(\frac{1}{a-1}+1\right) \\
= & (a-1)+\frac{1}{a-1}+2 .
\end{aligned}
$$

Given $a>1, b>1$, we know that when and only when $a-1=1$, i.e., $a=b=2$, $a+b$ reaches its minimum value of 4.