From Cubic Equation to Weierstrass Form
Start from the Hesse cubic: \(C_\lambda:\quad X_0^3+X_1^3+X_2^3 -3\lambda X_0X_1X_2=0.\)
Choose a flex point as the origin:
\(O=[1
0].\)
The tangent line at $O$ is \(X_0+X_1+\lambda X_2=0.\)
Now choose new projective coordinates so that \(O=[0:1:0], \qquad \text{tangent line at }O=\{Z=0\}.\)
One convenient choice is \(X=X_2,\qquad Y=X_0,\qquad Z=X_0+X_1+\lambda X_2.\)
Equivalently, \(X_0=Y,\qquad X_2=X,\qquad X_1=Z-Y-\lambda X.\)
Substitute this into the Hesse equation: \(Y^3+(Z-Y-\lambda X)^3+X^3 -3\lambda Y(Z-Y-\lambda X)X=0.\)
Because $O=[0:1:0]$ is now the point at infinity and $Z=0$ is its tangent line, this cubic can be written in generalized Weierstrass form: \(Y^2Z+a_1XYZ+a_3YZ^2 = X^3+a_2X^2Z+a_4XZ^2+a_6Z^3.\)
The coefficients $a_i$ are functions of $\lambda$, hence functions of $\tau$.
On the affine chart $Z=1$, this is \(y^2+a_1xy+a_3y = x^3+a_2x^2+a_4x+a_6.\)
Complete the square: \(y^{\prime} = y+\frac{a_1x+a_3}{2}.\)
Then shift $x$ to remove the $x^2$ term: \(x^{\prime} = x+\frac{a_2+a_1^2/4}{3}.\)
This gives \((y^{\prime})^2=(x^{\prime})^3+A(\tau)x^{\prime}+B(\tau).\)
After homogenizing and rescaling $Y$, we get the short Weierstrass form: \(Y^2Z=4X^3-g_2(\tau)XZ^2-g_3(\tau)Z^3.\)
Thus the path is: \(\text{Hesse cubic from theta} \longrightarrow \text{choose origin and tangent} \longrightarrow \text{Weierstrass form}.\)