8. Given that the sum of the distance from any point on curve $C$ to point $A(0,0)$ and the line $x=4$ is equal to 5, for a given point $B(b, 0)$, there are exactly three pairs of distinct points on the curve that are symmetric with respect to point $B$, then the range of $b$ is $\qquad$
$8.2<b<4$

Analysis: Let a point on curve $C$ be $(p, q)$, then $\sqrt{p^{2}+q^{2}}+|p-4|=5$, simplifying to: when $p \geq 4$, $p=\frac{81-q^{2}}{18}$; when $p<4$, $p=\frac{q^{2}-1}{2}$; since both segments are parabolas and are symmetric about the $x$-axis, the remaining two pairs of symmetric points must be distributed on the two segments of the parabolas. Assuming the point on the left parabola is $A(x, y), x<4$, its symmetric point is $B(s, t)$. Then $s=2 b-x, t=-y$.

Since $2 x=y^{2}-1, 18(2 b-x)=81-y^{2}$, thus $x=\frac{9}{4} b-5$, from $-\frac{1}{2}<x<4$ we know, $2<b<4$;