In general, two conics in the plane intersect at most 4 points. Suppose three of those points are given as $A,B,C$. Then let $c\_1$ be the conic passing through those three points and $D\_1,E\_1$. Let $c\_2$ be the conic passing through those three points and $D\_2,E\_2$. How can the fourth intersection point of these two conics be constructed geometrically, with ruler and compass?



Geometric construction of the fourth intersection points of two conics
Based on [*Projective conic sections - constructions*](http://mathafou.free.fr/themes_en/conique7.html),
the crux of the construction is this:


* let two conics intersect in $A,B,C,D$.
* let any line through $A$ intersect the conics again in $M,M'$
* let any line through $B$ intersect the conics again in $N,N'$
* then $MN, M'N'$ and $CD$ are concurrent.


To show this, consider the hexagons $MACDBN and M'ACDBN'.$ Let $P=MA\cdot DB=M'A\cdot DB$ and
$Q=AC\cdot BN=AC\cdot BN'$. By Pascal's Theorem $CD\cdot MN$ and $CD\cdot M'N'$ are on line $PQ$, and the concurrency follows. In particular, $F=MN\cdot M'N'$ lies on $CD.$


For any $T\neq U$, let $UT\cdot UVWXY$ denote the other intersection $Z$ of the line $UT$ with the conic defined by the five points $U,V,W,X,Y$. There is a classic straightedge construction of $Z$ based on Pascal's Theorem which is described in Hatton's [*Projective Geometry* (pg 240, 133.A.ii)](https://archive.org/details/cu31924060184045/page/n255/mode/1up)


Putting it all together, the steps for the construction of $D$ are:


* $M=AT\cdot ABCD\_1E\_1$
* $M'=AT\cdot ABCD\_2E\_2$
* $N=BT\cdot BACD\_1E\_1$
* $N'=BT\cdot BACD\_2E\_2$
* $F=MN\cdot M'N'$ (as mentioned above, $F$ will lie on $CD$)
* $D=CF\cdot CABD\_1E\_1$


Note that the construction can be done with a straightedge only - no compass required!


