Write $X=x\circ\phi$. Since $X(-P)=X(P)$, it is an even rational function and therefore belongs to $k(x)$; write $X=p(x)/q(x)$ with coprime polynomials $p,q$. The pullback of a nonzero invariant differential is invariant, so for some $c\in k^\times$,
$$
\phi^*\left(\frac{dX}{2Y}\right)=c\frac{dx}{2y}.
$$
Since
$$
dX=\frac{p'q-pq'}{q^2},dx,
$$
we obtain
$$
Y=\frac{p'q-pq'}{cq^2},y.
$$
Thus
$$
\phi(x,y)=\left(\frac{p(x)}{q(x)},\frac{p'(x)q(x)-p(x)q'(x)}{c q(x)^2}y\right).
$$


Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 125 / 3 / b / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/iii/paper-125/3/b/solution