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This question was inspired by another one I asked myself these days Given a drawing of an ellipse is there any geometric construction we can do to find it's foci?

I think this is harder, I can't even find is axis or vertex.

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    $\begingroup$ Any parabola with any orientation? $\endgroup$
    – tomi
    Mar 15, 2015 at 1:37
  • $\begingroup$ yes the orientation don't matter $\endgroup$
    – onlyme
    Mar 15, 2015 at 2:09

2 Answers 2

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Let three tangents to a parabola form a triangle. Then Lambert's theorem states that the focus of the parabola lies on the circumcircle of the triangle.

So draw three tangents. Find the triangle that they form and construct the circumcircle.

Start again with two of the original tangents and a new tangent. Find the triangle they form and construct the circumcircle. The focus must be at one of the two intersections of these circles.

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  • $\begingroup$ how to draw perfect tangents without knowing the focus and the axis of simetry ? $\endgroup$
    – onlyme
    Mar 15, 2015 at 3:41
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    $\begingroup$ @onlyme Given a parabola $\mathcal{P}$ and any direction not parallel to its axis of symmetry, one can construct a tangent along that direction as follows: Let $\ell_1, \ell_2$ be a pair of lines along that direction which intersect $\mathcal{P}$ at points $a_1, b_1$ and $a_2, b_2$. Let $m_1$ and $m_2$ be mid-points of line segments $\overline{a_1b_1}$ and $\overline{a_2b_2}$. Construct a line through $m_1$ and $m_2$ and let it intersect $\mathcal{P}$ at $c$. The line through $c$ parallet to $\ell_1$ and $\ell_2$ will be a tangent of $\mathcal{P}$ along the desired direction! $\endgroup$ Mar 15, 2015 at 4:44
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As @AchilleHui mentions, the midpoints of two parallel chords lead to the point of tangency ($T$) with a third parallel line. Note that the line of midpoints is parallel to the axis of the parabola. By the reflection property of conics, the line of midpoints and the line $\overleftrightarrow{TF}$ make congruent angles with the tangent line.

enter image description here

So, two sets of parallel chords determine two points of tangency and two lines-of-midpoints, which determine two lines that meet at focus $F$. $\square$

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  • $\begingroup$ Wow that solves it even faster! And also the way I found the axis with the other method shown above was way more complex: give the foci I would make two more tangents at some points, draw the perpendicular lines to that tangents through the foci intersect theses lines and get two points on the line parallet to the directrix on the apex of the parabola. Thanks. $\endgroup$
    – onlyme
    Mar 15, 2015 at 19:52

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