Two-isogeny formula ($\varphi :E⟶E^{\prime },\mathrm{deg}\varphi =2$)

For $E:y^2=x(x^2+ax+b)$, set $b^{\prime }=a^2-4b$ and $E^{\prime }:Y^2=X(X^2-2aX+b^{\prime })$. The isogeny of elliptic curves with kernel $\{O,(0,0)\}$ is $\varphi (x,y)=(x+a+b/x,y(1-b/x^2))$. Its dual is $\hat{\varphi }(X,Y)=((X-2a+b^{\prime }/X)/4,Y(1-b^{\prime }/X^2)/8)$. The maps extend over their exceptional affine points and satisfy $\hat{\varphi }\varphi =[2]$.