One, (20 points) Given that $x$, $a$, and $b$ are real numbers, and satisfy $y=\frac{a x^{2}+8 x+b}{x^{2}+1}$ with a maximum value of 9 and a minimum value of 1. Find the values of $a$ and $b$.


Given that $x^{2}+1>0$, we can simplify and rearrange the given equation by eliminating the denominator, resulting in
$$
(y-a) x^{2}-8 x+(y-b)=0 .
$$

Since $x$ is a real number, we have
$$
\Delta=(-8)^{2}-4(y-a)(y-b) \geqslant 0 \text {, }
$$

which simplifies to $y^{2}-(a+b) y+(a b-16) \leqslant 0$.
Given that $1 \leqslant y \leqslant 9$, we have $(y-1)(y-9) \leqslant 0$, which simplifies to
$$
y^{2}-10 y+9 \leqslant 0 \text {. }
$$

By comparing the coefficients of equations (1) and (2), we get
$$
\left\{\begin{array}{l}
a+b=10, \\
a b-16=9 .
\end{array}\right.
$$

Therefore, $a=b=5$.