For
$$
E:y^2+y=x^3-4x+2,
$$
the negative of $(x,y)$ is $(x,-y-1)$. At $P=(0,1)$ the tangent slope is
$$
\lambda=\frac{3x(P)^2-4}{2y(P)+1}=-\frac43.
$$
The tangent is $y=1-4x/3$, so the addition formulas give
$$
x(2P)=\lambda^2-2x(P)=\frac{16}{9},
\qquad
y(2P)=-\lambda x(2P)-1-1=\frac{10}{27}.
$$
Thus
$$
\boxed{2P=\left(\frac{16}{9},\frac{10}{27}\right).}
$$
The line through $P=(0,1)$ and $Q=(-2,1)$ is $y=1$. Its third intersection has $x=2$, and reflection under $y\mapsto-y-1$ gives
$$
\boxed{P+Q=(2,-2).}
$$


Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 125 / 1 / b / i / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/iii/paper-125/1/b/i/solution