Geometric construction of the fourth intersection points of two conics 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?
 A: 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!
A: In affine coordinates where $A=(a,0)$, $B=(0,b)$, $C=(0,0)$, the two conics have the equations
$$p_1(x^2-ax)+q_1(y^2-by)+r_1xy=0$$
$$p_2(x^2-ax)+q_2(y^2-by)+r_2xy=0$$
So the fourth point of intersection $F$ satisfies
$$\frac{y}{x-a}=\frac{p_1 q_2-p_2 q_1}{q_1 r_2-q_2 r_1}$$
$$\frac{y-b}{x}=\frac{r_1 p_2-r_2 p_1}{p_1 q_2-p_2 q_1}$$
The left sides are naturally interpreted in terms of the angles $FAC$ and $FBC$, so this might lead to a nice construction.
Update:
Using Cramer's rule to solve for the $p$'s, $q$'s, $r$'s, we can take
$$p_1=\begin{vmatrix}
D_{1y}^{\,2}-bD_{1y} & D_{1x}D_{1y}\\
E_{1y}^{\,2}-bE_{1y} & E_{1x}E_{1y}\\ 
\end{vmatrix}$$
and similarly for $q_1, r_1, p_2, q_2, r_2$. So translating all this into geometry seems like it will primarily involve ten constructions of determinant calculations.
