Timeline for When do two lines and three points determine exactly two conics? Exactly four?
Current License: CC BY-SA 3.0
10 events
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| Apr 13, 2017 at 12:19 | history | edited | CommunityBot |
replaced http://math.stackexchange.com/ with https://math.stackexchange.com/
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| Sep 20, 2016 at 17:14 | vote | accept | Zsbán Ambrus | ||
| Sep 19, 2016 at 11:59 | comment | added | David E Speyer | So there's always 0 or 4, never 2? That's interesting! Reminds me of arxiv.org/abs/1211.7160 . | |
| S Sep 19, 2016 at 9:16 | history | suggested | Zsbán Ambrus | CC BY-SA 3.0 |
make clear that all four of the solutions are real
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| Sep 19, 2016 at 8:56 | comment | added | Zsbán Ambrus | Thanks, the method in that post indeed makes it clear that all four solutions are real in that case. I suggested an edit to the post to state that explicitly. | |
| Sep 19, 2016 at 8:55 | review | Suggested edits | |||
| S Sep 19, 2016 at 9:16 | |||||
| Sep 19, 2016 at 8:48 | comment | added | MvG | If you follow the answer from my linked post, it boils down to the fact that three points in the plane will lift to three pairs of points on a cone, from which one can form $2^3$ planes. Two planes lead to distinct conics if projected back to the plane, unless they are symmetric wrt. the drawing plane. But they come in symmetric pairs, resulting in the $2^3/2=4$ distinct conics in general. I could give more details for each of these steps, but I don't see any of them as particularly problematic. | |
| Sep 19, 2016 at 8:35 | comment | added | Zsbán Ambrus | Thank you, this answer is helpful, but I don't think it's complete. I understand that when the two lines separate any two of the three points, then there are no real solutions. But when the lines don't separate the points, will there always be four real solutions in the general case? If so, can you give me a pointer for how to prove that? | |
| Sep 19, 2016 at 7:43 | history | edited | MvG | CC BY-SA 3.0 |
Interactive demo
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| Sep 19, 2016 at 7:16 | history | answered | MvG | CC BY-SA 3.0 |