Possible reasons why the image of 0 in the modular parametrization would always be O for a family of quadratic twists of elliptic curves Let $E$ be an elliptic curve and $E_n$ be its quadratic twist by $n$.  Let $\phi_n: X_0(N_n) \to E_n$ be the normalized modular parametrization of $E_n$ ($\infty \to O$)
For some particular curves $E$, it seems (based on computing some examples in sage) to always be the case that $\phi_n(0) = O$ when the rank of $E_n$ is 0.  For the curves I'm looking at, the torsion subgroup is $\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$ for all of the $E_n$.  What are some possible explanations for this or strategies to go about proving it? 
Note: Edited to clarify the first comment. 
 A: Note: this gives a plausible explanation, not a whole solution.
Hi, the BSD-quotient is $L(E_n,1)/\Omega_n = Sha\cdot\prod_p c_p(E_n)/|T|^2$. I think your question is mostly equivalent to asking why this quotient is an integer when twisting by $n$? The torsion is of size 4 (maybe there is an exceptional twist where it is larger, you did not specify). I think because of the full 2-torsion, the Tamagawa number should be 4 at any prime dividing $n$, except those dividing the original conductor. So if $n$ has at least two prime factors coprime to $N$ it seems you are done. When twisting by a prime or some $n$ with nontrivial gcd with $N$, then you need to take a more careful analysis, but the original curve itself should already have nontrivial Tamagawa numbers at some prime if indeed it has full 2-torsion. In short, the product of Tamagawa numbers outweighs the torsion in the denominator.
The definition of "0" here is just the cusp 0 on the boundary of the complex upper half plane. One is essentially integrating the modular form as $\int_0^{i\infty} f(z) dz$, to get the modular parametrization. But there are the constants like $2\pi$ and the Manin constant to worry about.
A: I only know this material in a vague and impressionistic way, but I think that it the fact that $0$ goes to the identity on the elliptic curve follows from Eichler-Shimura theory. 
If $E/\mathbb{Q}$ is parametrized by a modular curve, then it's a quotient of the Jacobian of the modular curve. In the composite map $$X_{0}(N) \to J_{0}(N) \to E$$ each arrow sends $0$ to $0$.
[Edit: Looking at this question I see that apparently 0 need not go to the identity in general. So now I'm puzzled. I think that my confusion might come from not knowing the definition of "0" on the modular curve in this context, which I was taking to be the preimage of the identity on the Jacobian.]
