32. Let $A$ and $B$ be points that lie on the parabola $y=x^{2}$ such that both are at a distance of $8 \sqrt{2}$ units from the line $y=-x-4$. Find the square of the distance between $A$ and $B$.
32. Answer: 98 .
Since $y=x^{2}$ and $y=-x-4$ do not intersect, $A$ and $B$ must lie on a line parallel to $y=-x-4$, namely, $y=-x+c$. The distance from $(0,-4)$ to $y=-x+c$ is
$$
8 \sqrt{2}=\frac{|(0)+(-4)-c|}{\sqrt{1+1}} .
$$

So $c=12$ or $c=-20$ (ignored). Substitute $y=-x+12$ into the parabola: $x^{2}=-x+12 \Rightarrow$ $x=3,-4$. So $A$ is $(3,9)$ and $B$ is $(-4,16)$. Then
$$
|A B|^{2}=(3-(-4))^{2}+(9-16)^{2}=98 .
$$