2. If $a b \neq 1$, and $5 a^{2}+2001 a+9=0$ and $9 b^{2}+2001 b+5=0$, then the value of $\frac{a}{b}$ is ( ).
(A) $\frac{9}{5}$
(B) $\frac{5}{9}$
(C) $-\frac{2001}{5}$
(D) $-\frac{2001}{9}$
2. (A).
$5 a^{2}+2001 a+9=0$,
$9 b^{2}+2001 b+5=0$ (obviously $b=0$ is not a solution of the equation)
$$
\Rightarrow 5 \cdot \frac{1}{b^{2}}+2001 \cdot \frac{1}{b}+9=0 \text {. }
$$

Therefore, $a$ and $\frac{1}{b}$ are both roots of the equation $5 x^{2}+2001 x+9=0$, but $a \neq \frac{1}{b}$. Given $\Delta>0$, that is, $a$ and $\frac{1}{b}$ are distinct real roots of this equation. Thus, $a \cdot \frac{1}{b}=\frac{9}{5}$.