Intersect point of two conic sections.
To construct with straightedge alone the 4th intersection point of two conic sections Γ and Γ'
Given two conic sections intersecting in A,B,C,D.
A line through A intersects the conic sections in two other points M,M'
A line through B intersects the conic sections in two other points N,N'

MN, M'N' and CD are concurrent.

In the following applet, the conic sections are defined by 5 points, the 4 points ABCD
and point E for Γ, and point F for Γ'.
The draggable green and cyan points define the intersecting lines AMM' and BNN'.

Proof

Consider Pascal theorem, with the two hexagons MACDBN and M'ACDBN'
Hence intersect points I = MA.DB, J = AC.BN and O = CD.MN are in line, hence O is intersect of IJ and CD.
Using Γ' : intersect points I' = M'A.DB, J' = AC.BN' and O' = CD.M'N' are in line, hence O' is intersect of I'J' and CD.
As line MA is M'A, I = I', and similarily J = J', hence O = O'.