Past exam of the mathematics course of the University of Cambridge / 2002 / iii / Paper 24 / 1 / b

Here the inverse in the elliptic curve group law is $(x,y)\mapsto (x,-y-1)$. For $P_0+P_1$, the line is $y=2-2x$. Substitution in the Weierstrass equation of an elliptic curve givesThe third intersection is $(3,-4)$, so reflection givesFor the difference, first take $-P_1=(1,-1)$. The line through $P_0$ and $-P_1$ is $y=2-3x$, and now the intersection polynomial is $x(x-1)(x-8)$. Its third point is $(8,-22)$, givingFinally, implicit differentiation gives $(2y+1)y^{\prime }=3x^2-7$. The tangent line at $P_1$ therefore has slope $-4$ and equation $y=4-4x$. Its intersection polynomial factors as $(x-1{)}^2(x-14)$. The double root records the intersection multiplicity of the tangent, and the remaining intersection is $(14,-52)$. Hence