Example 8 Given $a>0, b>0$, and $\sqrt{a}(\sqrt{a}+4 \sqrt{b})=3 \sqrt{b}(\sqrt{a}+2 \sqrt{b})$. Then the value of $\frac{a+6 \sqrt{a b}-8 b}{2 a-3 \sqrt{a b}+2 b}$ is ( ).
(A) 1
(B) 2
(C) $\frac{19}{11}$
(D) $\sqrt{2}$
(2007, Shandong Province Junior High School Mathematics Competition)
From the given, we easily obtain
$$
a+\sqrt{a b}-6 b=0 \text {. }
$$

Thus, $(\sqrt{a}-2 \sqrt{b})(\sqrt{a}+3 \sqrt{b})=0$.
Since $a>0, b>0$, then $\sqrt{a}+3 \sqrt{b} \neq 0$.
Therefore, $\sqrt{a}-2 \sqrt{b}=0 \Rightarrow a=4 b$. Hence, $\frac{a+6 \sqrt{a b}-8 b}{2 a-3 \sqrt{a b}+2 b}=2$.
So, the answer is (B).