TITLE: Geometric meaning of fiber of modular parameterization over a point of an elliptic curve?
QUESTION [16 upvotes]: Given an elliptic curve $E/\mathbb{Q}$ of conductor $N$, parameterization $\psi : X_0(N) \rightarrow E$, and a point $P \in E$, take the fiber $\psi^{-1}(P)$. Its points, being on $X_0(N)$, correspond to equivalence classes of pairs $(E_x, C_x)$.
Is there a geometric meaning for these pairs in relation to the point $P$? Something about these elliptic curves that has something to do with $P$? (Other than the j-invariant solving some polynomial equation with coefficients depending only on $P$ (and $E$ of course))
Anything special about the $C_x$'s in relation to $P$?
Modular parameterization is fascinating, but I just don't understand where it comes from. $X_0$ is for a whole bunch of elliptic curves. $E$ is a single specific one.What's the connection between points on the modular curve, to a single specific point of $E$?

REPLY [6 votes]: If your elliptic curve $E$ has rank $1$ over $\mathbb{Q}$, then there is a point $P$, a Heegner point, in $E(\mathbb{Q})$ with a special point $x=(E_x,C_x)$ in the fibre of $\psi$, i.e. such that $E_x$ has extra endomorphisms (as described in the other answers).
If instead your point $P$ of infinite order is on a curve $E$ of rank $2$ over $\mathbb{Q}$, then I believe there is nothing known about the points in the fibre. The degree $d$ of $\psi$ is getting large quite quickly. Each of the $d$ points will be defined over a large number field $K_x$. Probably in most cases the points in the fibre will form one single Galois-orbit. I believe this was verified for the generators of the curve of conductor 389. And there did not seem anything special about the points in the fibre.
It is a strange coincidence that elliptic curves over $\mathbb{Q}$ admit a cover by a curve which parametrises elliptic curves with extra structure. 
Given that there is this coincidence, one can ask what happens if one takes the points $x=(E,C)$ for different $C$'s on $X_0(N)$ and map them to $E$. This point $P=\psi(x)$ will not be defined over $\mathbb{Q}$ unless $N=27$. (There are two cm curves of conductor 27 with a 27 isogeny between them; the point $P$ is torsion in this case.) Otherwise, if the $j$-invariant of $E$ is not in $\tfrac{1}{2}\mathbb{Z}$ then $P$  will be of infinite order, but it will be defined over a number field much larger than $\mathbb{Q}$.