TITLE: Geometric construction of the fourth intersection points of two conics
QUESTION [10 upvotes]: 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?

REPLY [5 votes]: Based on Projective conic sections - constructions,
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)
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!