2. For the parabola $y=a x^{2}+b x+1$, the parameters $a$ and $b$ satisfy $8 a^{2}+4 a b=b^{3}$. Then, as $a$ and $b$ vary, the vertex of the parabola must lie on ( ).
(A) parabola
(B) hyperbola
(C) circle or ellipse
(D) line
2.B.

The vertex coordinates of the parabola $y=a x^{2}+b x+1$ are $\left(-\frac{b}{2 a}, \frac{4 a-b^{2}}{4 a}\right)$. Let $x=-\frac{b}{2 a}, y=\frac{4 a-b^{2}}{4 a}$. Then $\frac{b}{a}=-2 x, y=1-\frac{b^{2}}{4 a}=1+\frac{b x}{2}$.

Since $a \neq 0$, the conditions that $a$ and $b$ satisfy are equivalent to $8+4 \cdot \frac{b}{a}=b\left(\frac{b}{a}\right)^{2}$. Therefore,
$$
8+4(-2 x)=b(-2 x)^{2}=4 x(2 y-2) \text {, }
$$

which means $x y=1$.