It is an easy exercise that using ruler and compass one find the focus of a given parabola.

Can one do the same using only a ruler? -- if not, why?

  • $\begingroup$ Is this for a ruler or a straightedge and compass? $\endgroup$ – Random832 Dec 31 '14 at 20:22
  • $\begingroup$ @Random832: yes, perhaps, but i don't know -- my native languages are Russian, Maths and "things-as-they-are". But if you talk about en.wikipedia.org/wiki/Straight_edge then i agree! $\endgroup$ – Victor Jan 1 '15 at 6:13

No, because there exists a projective map which preserves the parabola but does not preserve its focus (this is so for any conic and any point.)

  • $\begingroup$ Fantastic! It's very easy and hence beautiful and hence genius! Happy New Year $\endgroup$ – Victor Dec 31 '14 at 17:38
  • $\begingroup$ How do you prove this? $\endgroup$ – Ivan Meir Apr 28 '16 at 7:32
  • $\begingroup$ You need to define the focus and by doing so I guess you might already see that this is not projective. Also there are points that are always preserved by a projective transform, the point at infinity in the case of the parabola so this contradicts your final statement. Can you modify it to make it always hold? $\endgroup$ – Ivan Meir Apr 28 '16 at 7:39
  • $\begingroup$ Assume that there is some construction on the plane $\pi$ which given parabola $\gamma$ gives its focus $F$. Let $T$ be projective map for which $T(\gamma)$ is a parabola again, but $T(F)$ is not it's focus. Apply the same construction to $T(\gamma)$ on $T(\pi)$, we get $T(F)$. Contradiction. $\endgroup$ – Fedor Petrov Apr 28 '16 at 8:35

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