Suppose I have a non-singular elliptic curve:
$$ y^2 = x^3 + ax^2 + bx +c$$
From the graph I know it either has 1 or 3 real roots, and that the slope of the tangent line at those points is $\infty$, but I am not sure how to do this without appealing to the graph.
I know if we let $F = y^2 – x^3 + ax^2 + bx +c$, then $$\frac {\delta F}{\delta y} = 2y \text{ and } \frac {\delta F}{\delta x} = 3x^2 + 2ax +b $$
So $ $$\frac {\delta F}{\delta y} (a,0) = 0$, but I don't know what this tells me that the slope of the tangent line at $(a,0)$.
Best Answer
An elliptic curve is symmetric around the $x$ axis, if $(x,y)$ lies on it, then so does $(x, -y)$. Can you prove from this algebraically that a tangent line through a point with $y=0$ has infinite slope?
Edit: Consider $(x(t),y(t))$ a parametrization of a small piece of the elliptic curve such that $y(0)=0$. Then by the symmetry $(x(t),-y(t))$ is also a parametrization of a small piece of the elliptic curve that goes through the same point $(x(0),0)$ in $\mathbb{R}^2$.
Now there are two possibilities, either the two parametrizations represent the same part of the elliptic curve, then $y'(0)=0$ or they represent different segments of the elliptic curve, then $(x(0),0)$ is a double point with two distinct tangent lines.