Embedding of elliptic curves into $\mathbb{P}^2$ by arbitrary line bundle of degree $3$ Let $E$ be a complex elliptic curve, with distinguished point $x_0 \in E$. Any divisor of degree three is equivalent to the divisor $D=x+2x_0$. If $x=x_0$, it is well known that $L(D)$ has an explicit basis that is usually used as the standard embedding to projective plane 
$$L(D) \cong span(1, \wp, \wp'),$$ where $\wp$ is the Weierstrass function. 
What functions can I choose as a basis for an arbitrary $D$ of degree $3$? 
Of course, the Weierstrass function and constants still belong to $L(D)$, but what is the third function?
 A: AFAICT the answer depends on $x$ a little.
If $x$ is one of the half-periods, then $\wp'$ has a simple zero at $x$,
the other simple zeros of $\wp'$ being the other two non-trivial half-periods, call them $x_2$ and $x_3$. Then the function
$$
f=\frac{(\wp-\wp(x_2))(\wp-\wp(x_3))}{\wp'}
$$
has a simple pole at $x_0$, and a simple pole at $x$. The other zeros of the denominator $\wp'$ are cancelled by the factors in the numerator. The pole at $x_0$ is simple, because the pole order of $\wp$ at $x_0$ is two, and that of $\wp'$ is three. 
OTOH, if $x$ is not one of the half-periods, then the points $\pm x$ are distinct. It is known that $\wp(z)=\wp(x)$, iff $z$ is congruent to $\pm x$
modulo the lattice of periods. Therefore the function
$$
f=\frac{\wp'-\wp'(-x)}{\wp-\wp(x)}
$$
has no pole $-x$, because the zero of the denominator at $-x$ is simple. This function thus has a simple pole at both $x_0$ and $x$, and can be used.
It may be easier to think about this in terms of the Weierstrass form of the elliptic curve, and the points $P(x)=(\wp(x),\wp'(x))=(u,v)$ that lie on the curve $$v^2=4u^3-g_2u-g_3.\qquad(*)$$
In the case of a half-period, the point $P(x)$ on the curve $(*)$ has a vertical tangent. In the other cases $\wp-\wp(x)$ is a local parameter at $x$, and we simply use the factor $\wp'-\wp'(-x)$ to cancel the pole at the other zero of the denominator. A cleaner way of using this may be out there.
