5. A. Given that $a, b$ are non-zero real numbers, on the parabola $y=a x^{2}+b$ there are three different points $A, B, C$, and it is known that $B C \parallel x$-axis, $\angle B A C=90^{\circ}$. Then the distance from point $A$ to the line $B C$ is ( ).
(A) $\frac{1}{|a|}$
(B) $\frac{1}{2|a|}$
(C) $|a|$
(D) $2|a|$
5. A. A.

Since the parabola $y=a x^{2}+b$ is symmetric about the $y$-axis, and $B C / / x$-axis, therefore, points $B$ and $C$ are symmetric about the $y$-axis.

Let $A\left(s, a s^{2}+b\right), B\left(-t, a t^{2}+b\right), C\left(t, a t^{2}+b\right)$, where $t>0, s \neq \pm t$.
Then, from $\angle B A C=90^{\circ}$, we get
$$
\begin{array}{l} 
A B^{2}+A C^{2}=B C^{2} \\
\Rightarrow(s+t)^{2}+\left[a\left(s^{2}-t^{2}\right)\right]^{2}+(s-t)^{2}+ \\
{\left[a\left(s^{2}-t^{2}\right)\right]^{2}=(2 t)^{2} } \\
\Rightarrow a^{2}\left(s^{2}-t^{2}\right)=-1 .
\end{array}
$$

Thus, the distance from point $A$ to the line $B C$ is
$$
\begin{aligned}
h & =\left|\left(a s^{2}+b\right)-\left(a t^{2}+b\right)\right| \\
& =\left|a\left(s^{2}-t^{2}\right)\right|=\frac{1}{|a|} .
\end{aligned}
$$