If two parabolas don't intersect, is there a line that doesn't intersect either of them?

Question

Two parabolas in a plane are given, such that they don't intersect. Is it true that there is a line in plane such that doesn't intersect any of them?

Answer 1

Parabolas are convex, so they'll always be on the same side of their tangent lines. Find both tangent lines at the points of closest approach. These lines are parallel. Intuitively, if they were not then taking a 'small step' along the parabola in the right direction would bring you closer. (Slightly) more rigorously, if you know how to derive the method of lagrangian multipliers, this is a similar idea. so you can pick any line between these parallel lines to satisfy the problem.

If one is contained inside the other the above won't work, because even when a point of closest approach does exist, the parabolas will both be on the same side of the tangent line. It's a simple case though, just use the directrix of the larger parabola as your answer.

Answer 2

To take a somewhat more high-brow approach: Consider the parabolas together with their interiors (in the obvious sense, e.g., the interior of $y=x^2$ is $y>x^2$). Now you have two closed, convex sets whose boundaries don't intersect. Either one is contained within the other, or the two sets are disjoint. The former case is trivial. In the latter case, use the Hahn–Banach separation theorem to find a line (really, a level set of a linear functional) separating the two.

Edit: Due to lack of compactness [see the comments], the above method might yield a common tangent line to the two parabolas. To give ourselves some wiggle room, we can use the following “compact slices” property of the convex hull $C$ of a parabola (easily proved in the standard configuration $y=x^2$): If $f(C)$ is bounded below where $f$ is a linear functional, then $C\cap f^{-1}((-\infty,m])$ is compact for any $m$.

So let $C_1$, $C_2$ be disjoint convex hulls of parabolas. From Hahn–Banach, we get a linear functional $f$ with $\sup f(C_1)\le m:=\inf f(C_2)$. Using the compactness of the slice $C_2\cap f^{-1}((-\infty,m+1])$, we can translate $C_2$ a small distance along its axis of symmetry to get $C_2'$ disjoint from $C_1$ and containing $C_1$ in its interior. Now apply Hahn–Banach once more to get a linear functional $g$ with $\sup g(C_1)\le\inf g(C_2')$. Since $\inf g(C_2')\lt\inf g(C_2)$, a level set of $g$ solves the problem.

Answer 3

Here's a simple approach, which I hope will suit.

Here is a parabola, courtesy of WolframAlpha:

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Here is another parabola.

Do the parabolas intersect? (No.) Is there a line, say y = -0.25, that won't intersect either parabola? (Yes, and there are others.)