5. Let the equation of a parabola be $y=a x^{2}+b x+2(a \neq 0)$, and it is known that for any real number $x$, $|y|$ $\geq 2$. Then the coordinates of the focus of this parabola are ( ).
A. $\left(0, \frac{1}{4 a}+2\right)$
B. $\left(0,2-\frac{1}{4 a}\right)$
C. $(0,0)$
D. $\left(0, \frac{b}{2 a}-2\right)$
5. A

When $x=0$, $y=2$. Since for any real number $x$, there is always $|y| \geqslant 2$, it follows that for any real number $x$, there is always $y \geqslant 2$, and the equality can be achieved. Therefore, $x=0$ is the axis of symmetry of the parabola $y=a x^{2}+b x+2$, and $a>0$, i.e., $y=a x^{2}+2(a>0)$. Thus, the focus of the parabola $x^{2}=\frac{1}{a}(y-2)=2 \cdot \frac{1}{2 a}(y-2)$ is $\left(0, \frac{1}{4 a}+2\right)$.