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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?

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Do you mean parabolas as graphs of functions $y=f(x) = ax^2 + bx + c$? Or are we talking about general parabolas? – Thomas Jun 7 '12 at 15:17
Several errors still remain at this time in the answer by Robert Mastragostino. – Michael Hardy Jun 7 '12 at 16:11
From a certain point of view, a parabola is an ellipse that has the line at infinity as one of its tangent lines. For the parabola $y=x^2$, if we pick a different line to serve as the line at infinity, namely $y=-1$, then the parabola becomes an ellipse. So if two parabolas don't intersect, the question is whether there is some line we could choose to regard as being at infinity, such that both parabolas become ellipses. If two ellipses in the projective plane don't intersect, we can always find a line far far away from both of them. – Michael Hardy Jun 7 '12 at 16:19

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.

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And the points of closest approach exist why? – Phira Jun 7 '12 at 14:59
Could you clarify a bit on why the parabolas are always outside of the cone? Also, the cone might not be cone, but instead just two parallel lines. – Zev Chonoles Jun 7 '12 at 14:59
@Phira: It's a standard argument about why there is always two points achieving $d(F,G)$ when $F,G$ are closed subsets of $\mathbb{R}^n$ (use compactness). – Najib Idrissi Jun 7 '12 at 15:02
@N.I: There need not exist such points if neither $F$ nor $G$ is compact. Consider the graphs of hyperbolas $xy = c$ for different values of $c$: $d(F,G) = 0$ but these sets have no points in common. – Noah Stein Jun 7 '12 at 15:11
This answer is wrong. Look at $y=x^2$ and $y=x^2+1$. They don't intersect; no point is on both curves. There is indeed a line that intersects neither of them; for example $y=-1$. If you pick any tangent line to $y=x^2+1$, it intersects $y=x^2$, and so does every line between that tangent line and the tangent to the other parabola that is parallel to it. – Michael Hardy Jun 7 '12 at 15:46

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.

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If I recall correctly, there's something non-constructive in the proof of the Hahn-Banach theorem. But here you should be able to get by with a much weaker version that is fully constructive. You should be able to actually find the line. – Michael Hardy Jun 7 '12 at 15:49
@MichaelHardy: Yes, and yes. Left as an exercise for the reader. ☺ – Harald Hanche-Olsen Jun 7 '12 at 16:36
Doesn't Hahn-Banach require compactness? – Grumpy Parsnip Jun 7 '12 at 22:21
@JimConant: There are many variants of Hahn–Banach separation. Not all require compactness. But without it, you may not get strict separation, and that is a problem for my proof. I have edited the answer to deal with this issue. Thanks for pointing it out. – Harald Hanche-Olsen Jun 8 '12 at 8:37
@MichaelHardy: To expand a little on my first “yes” above, the non-constructiveness in the proof of Hahn–Banach only comes into play in infinite dimensional spaces. After all, one expands a linear functional by one dimension at the time, which only requires ordinary induction in the finite-dimensional case (and not even that in the two-dimensional case, of course). – Harald Hanche-Olsen Jun 8 '12 at 8:39

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

Here is a parabola, courtesy of WolframAlpha: parabola facing up

Here is another parabola. parabola facing down

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

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This is not at all a complete answer to this question. -1 – mixedmath Jun 7 '12 at 19:55
Responding to an answer by saying it is incomplete without saying why is not helpful. The reason this answer is incomplete is because it doesn't show that for ANY two parabolas that do not intersect there exists a line that does not intersect either. Just that for the two parabolas you chose. – john w. Jun 7 '12 at 21:55
This works only when the parabolas are situated like the two in your example. In particular, their axes are parallel to each other. – Michael Hardy Jun 8 '12 at 11:40
Yes, I agree that I have provided a limited case. From the tags 'homework' and 'algebra-precalculus' I was guessing that the OP was seeking an example instead of a rigorous proof, and would appreciate visual insight over the Hahn-Banach theorem. This depends on whether the OP is in tenth grade precalculus or in a sophomore college calculus course. (I looked at the OP's profile, but his or her age was not revealed.) – rajah9 Jun 8 '12 at 13:35
Well, in that way you can prove that multiplication is the same as addition, thus: $2+2=2\cdot2$ and $1+2+3=1\cdot2\cdot3$. – Michael Hardy Jun 8 '12 at 17:56

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