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On this site it says, that we can make a conic with five tangents. I know how to construct a conic with five points, so I am wondering, how do we determine which point from tangents do we use. I know, that we cannot take three collinear points (so we cannot choose the intersections), but other than that, what are our restraints?

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  • $\begingroup$ I can't seem to find my copy, but Dorrie is good on specific constructions store.doverpublications.com/0486613488.html $\endgroup$ – Will Jagy Aug 23 '19 at 0:30
  • $\begingroup$ The contruction in Dorrie is online at archive.org/details/… $\endgroup$ – brainjam Aug 23 '19 at 4:19
  • $\begingroup$ Working algebraically, you can construct the matrix of the dual conic directly from the lines and then compute its adjoint to get the conic. The elements of $\mathbb R^3$ that represent the lines are therefore subject to the same constraints that five points would be. So, for instance, no three of the lines can be coincident. $\endgroup$ – amd Aug 23 '19 at 21:57
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Use Brianchon's theorem as in the following diagram:

enter image description here

If $F$ is the point of contact of the tangent $CD$, then by Brianchon's theorem the three diagonals of the hexagon $ABCFDE$ are concurrent at a point $O$.

So we can intersect $BD$ and $CE$ to find $O$, and the contact point $F$ lies on the line $AO$. Similarly we can find the four other contact points.

This gives us five points, from which we can construct the conic.

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