Example 3 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 $a c=$ $\qquad$
(2003, National Junior High School Mathematics League)
【Analysis】This is a comprehensive problem of algebra and geometry. The double perpendicularity in the problem is the key to solving it.

Solution Let $x_{1} 、 x_{2}\left(x_{1}<x_{2}\right)$ be the two zeros of the parabola. Then
$$
\begin{array}{l}
y=a x^{2}+b x+c=a\left(x-x_{1}\right)\left(x-x_{2}\right) \\
\Rightarrow a x_{1} x_{2}=c .
\end{array}
$$

Since $\triangle A B C$ is a right triangle, we have
$$
O A \cdot O B=O C^{2} \Rightarrow-x_{1} x_{2}=c^{2} \Rightarrow-a c^{2}=c \text {. }
$$

Since $c \neq 0$, it follows that $a c=-1$.