Let $E = V_+(F(u,v,w)) \subset \mathbb{P}^2_k$ be an elliptic curve. Let $o = (0,1,0)$ be the origin and $x \in E(k)$ a rational point. Let us suppose there is a curve $C \subset \mathbb{P}^2_k$ such that $C \cap E = \{x \}$. Show that $x-o \in Pic^0(X)$ is of finite order.
I haven't been able to kill this off, my thinking was as follows: We want to use the defining equation of C to use it as a divisor on E. However, we should then divide by a high enough power of some element in E to get it to be principal, I guess. But I am not sure about how to show that this is a principal divisor, or how to pull-back my divisor.
First, we see that $0$ is a flex point of $E$, since after a change of coordinates we can arrange $(0,1,0)$ to be the point "at infinity", and so the line "at infinity" $\{z=0\}$ intersects $E$ only at $(0,1,0)$ and with multiplicity 3.
Let $C$ be a curve of degree $d$; then by Bezout we have that the intersection (as divisors) of $C$ with $E$ is $3dx$. If $C$ is defined by the homogeneous polynomial $f$, then $f/z^d$ is a rational function on $\mathbb{P}^2$ such that $\mbox{div}((f/z^d)|_E)=3dx-3d\{0\}$ (I'm writing $\{0\}$ so as not to confuse the point 0 with the prime divisor whose support is the origin), and so $3d(x-\{0\})=0$ in $\mbox{Pic}^0(E)$.