12. 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 trajectory equation of the vertex $(s, t)$ of the parabola is $\qquad$
12. $s t=1$.

Notice that, the coordinates of the vertex 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 $s=-\frac{b}{2 a}, t=\frac{4 a-b^{2}}{4 a}$. Then
$$
\frac{b}{a}=-2 s, t=1-\frac{b^{2}}{4 a}=1+\frac{b s}{2} \text {. }
$$

Since $a \neq 0$, the conditions that $a, b$ satisfy are equivalent to
$$
\begin{array}{l}
8+4 \frac{b}{a}=b\left(\frac{b}{a}\right)^{2} \\
\Rightarrow 8+4(-2 s)=b(-2 s)^{2}=4 s(2 t-2) \\
\Rightarrow s t=1 .
\end{array}
$$