# Chosing points so that $5$ tangents determine a conic

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?

• I can't seem to find my copy, but Dorrie is good on specific constructions store.doverpublications.com/0486613488.html – Will Jagy Aug 23 '19 at 0:30
• The contruction in Dorrie is online at archive.org/details/… – brainjam Aug 23 '19 at 4:19
• 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. – amd Aug 23 '19 at 21:57

## 1 Answer

Use Brianchon's theorem as in the following diagram:

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.