2. The parabola $y=a x^{2}+b x+c$ intersects the $x$-axis at points $A$ and $B$, and the $y$-axis at point $C$. If $\triangle A B C$ is a right triangle, then $a c=$ $\qquad$
2. -1 .

Let $A\left(x_{1}, 0\right)$ and $B\left(x_{2}, 0\right)$. Since $\triangle ABC$ is a right triangle, it is known that $x_{1}$ and $x_{2}$ must have opposite signs, so $x_{1} x_{2}=\frac{c}{a}<0$. By the projection theorem, we know that $|O C|^{2}=|A O| \cdot|B O|$, which means $c^{2}=\left|x_{1}\right| \cdot\left|x_{2}\right|=\left|\frac{c}{a}\right|$.

Therefore, $|a c|=1, a c=-1$.