Flex coordinates for a smooth plane cubic
= Flex coordinates for a smooth plane cubic
{title2=$F=aX^3+Z(bY^2+cXY+dX^2)+Z^2(eY+fX)+gZ^3$}
Put a <flex> at $[0:1:0]$ with <tangent line> $Z=0$. The tangent section has a triple zero there, so $F(X,Y,0)=aX^3$ with $a\ne0$. Smoothness gives $b\ne0$. In characteristic different from two, replacing $Y$ by $Y+(cX+eZ)/(2b)$ completes the square and produces a <Weierstrass equation of an elliptic curve> using only projective linear coordinate changes.