TITLE: Is there a geometric construction that finds the intersection of two conics, given only five points on each?
QUESTION [7 upvotes]: The question: Five points define a conic, so let points $A,B,C,D,E$ and $F,G,H,I,J$ define two conics $c$ and $d$.  Is there a geometric construction that identifies the common points of $c$ and $d$?
(It's easy enough to do it in Geogebra, or algebraically.  What I'm asking for here is a ruler and compass construction.)
Background and context  Given, say, an ellipse $c$ it is not too difficult to construct tangents from a point $P$ to $c$ using just a straightedge.  But this assumes that the ellipse has been drawn as a given.  If only the points $A,B,C,D,E$ on $c$ have been given, you can easily construct the polar of $P$ but then have to construct the intersection of the polar and the ellipse.  This is less straightforward but the construction is described in several 19th century projective geometry texts (e.g. Cremona or Russell),and requires a reference conic (or compass) in addition to a straightedge.
What I haven't been able to find is a construction for the intersection of two ellipses/conics that are not pre-drawn but given only as two sets of five points each.  Hence my question.  It came up in following a certain construction for "imaginary" chords that involved finding the common points of two conical loci (Russell, XXVII 6.).
Update: Given two or three points of intersection it is possible to construct the remaining points of intersection.
Update 2: This problem is discussed in Veblen and Young, Projective Geometry, Vol I, $\S102$.

REPLY [2 votes]: As a simpler counterexample, consider the parabola $y=x^2$ and the circle $y^2=x(1-x)$.  Both have infinitely many constructible points and they intersect at the origin.  But the second point of intersection requires solving $x^4=x(1-x)$ with $x$ nonzero, leading to the irreducible equation $x^3+x-1=0$ whose real root cannot be (Euclidean) constructed.