Rays in fat parabolas

Question

Let $\epsilon>0$. Let $F\subseteq \mathbb{R}^2$ be the set of all points that lie at a distance less than $\epsilon$ from the curve $y=x^2$. Can $F$ contain a ray? That is, is there (for some $\epsilon$) a point $p\in F$ and non-zero $q\in \mathbb{R}^2$ such that all points of the form $p+qt$ for $t\ge 0$ lie in $F$? One might suspect that one could replace $y=x^2$ with any algebraic curve that has no branch at infinity asympotic to a ray, and the answer would be "no". But I don't see how to approach that.

Answer 1

I believe this is impossible. It's easier to consider lines instead of rays, and show that any line must diverge from the parabola in both directions. The simplest way to do this is probably via case analysis: break your lines into those of the form $x=c$ for some constant $c$, and then the usual $y=ax+b$ (WLOG you can take $a\geq 0$ here). For the former the result is trivial, since the parabola gets 'infinitely far away' in both $x$ directions; for the latter, you should be able to get bounds (in terms of $x$) on $y-x^2$ for any points $p=(x,y)\in F$ and use this characterization to show that the intersection with any particular line must be finite.