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To find the center of a circle, it's enough to choose three points on the circle and find the circumcenter of a triangle with those three points.

When a parabola is drawn and the formula is not given, how can I find its focus?

I add an example.

Can you find the center, vertex, axis, or directrix of the parabola?

A parabola was drawn.

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If I well understand the problem, I think that here there is a solution. Note that the key problem is to construct the axis, than you can use the answer in the related question cited in the comment of @Blue

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  • $\begingroup$ This uses more than the minimal four points (+ line at infinity) of the parabola, however it is probably quite easy to improve into an optimal solution that uses only four points (and only straightedge (Pascal's theorem on the six points on a conic) to build the oblique axis; then the construction becomes metric and we may use the linked solution). $\endgroup$ Apr 16, 2015 at 15:26
  • $\begingroup$ From OP we see that the parabola is "drawn"... so this seems a good solution. $\endgroup$ Apr 16, 2015 at 15:32
  • $\begingroup$ Thanks! That's what I want tot know. $\endgroup$
    – P.-S. Park
    Apr 16, 2015 at 16:14
  • $\begingroup$ I accept the downvotes, but I'm interesting to know if there is some wrong in the answer. $\endgroup$ Apr 16, 2015 at 18:29
  • $\begingroup$ Using a "drawn" figure in a compass-and-straightedge construction generally means using the minimal number of points defining it: namely two points for a line, or for a circle, either the center and one point, or three points. Using your solution plus Pascal's theorem, we can reduce this to: given a conic H through four points ABCD and tangent to a line L, and given any line X through A, how to find the second point of intersection of H and X. (For the parabola, L will be the line at infinity). Related questions: what about conics defined by 3 points + 2 tangents, etc ? $\endgroup$ Apr 17, 2015 at 10:40

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