 Solution

Work in characteristic different from two, as in the number-field application below. Nonsingularity is equivalent to $b(a^2-4b)\ne 0$. The chord through $P=(x,y)$ and $T=(0,0)$ has slope $y/x$. Using $y^2/x^2=x+a+b/x$ in the elliptic-curve addition formula givesThese formulas hold for $P\ne O,T$; addition interchanges $O$ and $T$.

It follows that $\xi =x+a+b/x=y^2/x^2$ and $\eta =y(1-b/x^2)$. The relation isFor a direct verification, observe thatTherefore the two-isogeny formula isThe target elliptic curve is nonsingular because its corresponding coefficient product is $16b(a^2-4b)\ne 0$. The rational map of projective varieties extends over the exceptional points to a morphism of smooth projective algebraic curves; at both $O$ and $T$ its affine coordinates tend to infinity, giving the displayed values. A nonconstant morphism between elliptic curves sending $O$ to $O$ is a group homomorphism, so this is an isogeny of elliptic curves.

One can also see the quotient directly: translation by $T$ leaves $\xi ,\eta$ invariant. The equation $x^2+(a-\xi )x+b=0$ makes the source function field a degree-two extension of the target function field; its nontrivial automorphism is translation by $T$. Equivalently, the degree of an isogeny from its x-coordinate map is two. The kernel of an isogeny is precisely $\{O,T\}$. Thus $\varphi$ is a separable isogeny of degree two.