7. The parabola $y=a x^{2}-b x+b$ intersects the line $y=a^{2} x$ (where $a>0$) at points $P$ and $Q$. The absolute value of the difference in the x-coordinates of $P$ and $Q$ is 1. Let $d$ be the maximum distance from a point on the parabolic arc $PQ$ to the line. (I) Express $d$ in terms of $a$; (II) For what values of $a$ and $b$ is $d$ maximized?
(1991 Japanese College Entrance Examination)
7. (I) Let the line $P Q$ form an angle $\theta$ with the $x$-axis, and the distance from a point on the parabola with abscissa $x$ to the line $P Q$ be $l(x)$. The length of the line segment between the line $P Q$ and the parabolic arc, perpendicular to the $x$-axis at the same point, is $h(x)$. Then, $l(x)=h(x) \cdot \cos \theta$, and $h(x)=a^{2} x-\left(a x^{2}-b x+b\right)$. Note that the absolute difference in the abscissas of $P$ and $Q$ is 1, so we can assume their abscissas are $\alpha \mp \frac{1}{2}$. Also, the abscissas of points $P$ and $Q$ are the solutions to $h(x)=0$, so $h(x)=-a\left[x-\left(\alpha-\frac{1}{2}\right)\right]\left[x-\left(\alpha+\frac{1}{2}\right)\right]=-a\left[(x-\alpha)^{2}-\frac{1}{4}\right]$. Therefore, when $x=\alpha$, $h(x)$ is maximized.
Furthermore, $\cos \theta=\frac{1}{\sqrt{a^{4}+1}}$, and $h(\alpha)=\frac{1}{4} a$, so $d=h(\alpha) \cdot \cos \theta=\frac{a}{4} / \sqrt{a^{4}+1}$.
(II) $d=\frac{1}{4} / \sqrt{a^{2}+\frac{1}{a^{2}}}$, and $a^{2}+\frac{1}{a^{2}} \geqslant 2$ (equality holds when $a^{2}=\frac{1}{a^{2}}$, i.e., $a=1$), so $d$ is maximized when $a=1$. At this time, let $h(x)=a^{2} x-\left(a x^{2}-b x+b\right)=0$, whose roots are $\frac{1}{2}(b+1) \pm \sqrt{(b+1)^{2}-4 b}$. Since the absolute difference between the roots is 1, we have $(b+1)^{2}-4 b=1$, solving which gives $b=0,2$. Therefore, $(a, b)=(1,0)$ or $(1,2)$ maximizes $d$.