Elliptic-curve addition formula

Let $E:y^2=x^3+a_2{x}^2+a_{4}x+a_6$ be a smooth Weierstrass equation of an elliptic curve over a field of characteristic different from two, with identity $O$ at infinity. For affine points $P,Q$ with $x(P)\ne x(Q)$, putFor $P=Q$ with $y(P)\ne 0$, use the tangent slope $\lambda =(3x(P{)}^2+2a_{2}x(P)+a_4)/(2y(P))$. In either case,Substitute the secant or tangent equation into the cubic: the sum of its three intersection $x$-coordinates is ${\lambda }^2-a_2$. Reflecting the third intersection in $y=0$ gives the sum.^[ref]

The omitted cases belong to the elliptic curve group law on the full projective curve: $P+O=P$ and $P+(-P)=O$, where $-P=(x(P),-y(P))$. In particular, a point with $y(P)=0$ satisfies $2P=O$; its vertical tangent cannot be handled by dividing by $2y(P)$.

On $y^2=x^3+ax+b$ in characteristic of a field other than two, take the elementary symmetric polynomials $s_i$ of $x(P),x(Q),x(P+Q)$. If the line through $P,Q$ is $y=\lambda x+\mu$, its third intersection is $-(P+Q)$, so coefficient comparison gives $s_1={\lambda }^2$, $s_2=a-2\lambda \mu$, $s_3={\mu }^2-b$. Hence $(s_2-a{)}^2=4s_1(s_3+b)$. This is an identity of rational functions; tangent cases follow by specialization, and coordinates at infinity are handled through the projective addition map.