TITLE: Show that the curve $y^2 = x^3 + 2x^2$ has a double point, and find all rational points
QUESTION [5 upvotes]: Show that the curve $y^2 = x^3 + 2x^2$ has a double point. Find all rational points on this curve.
By implicit differentiation of $x$, $-3x^2 - 4x$ vanishes iff $x = -4/3$ and $0$.
By implicit differentiation of $y$, $2y$ vanishes iff $y = 0$.
Taking the second derivative, I got $-6x-4$ and then using the point on the curve $(0,0)$ I got $-4$. Is this my double point?
Thanks for any help!

REPLY [2 votes]: You may find the following fact useful: suppose that a plane curve is described by a polynomial equation $F(x,y)=0$. Write
$$
F(x,y)=\sum_{d\geq0}F_d(x,y)\qquad\qquad(*)
$$
where in $F_d(x,y)$ you collect all the monomials in $F$ of degree exactly $d$. Then:

$(0,0)$ belongs to the curve if and only if $F_0\equiv0$ (by which, I mean that is $0$ identically).
$(0,0)$ is a singular point of the curve if and only if $F_0\equiv0$ and $F_1\equiv0$.

Moreover if $d_0\geq2$ is the smallest degree appearing in $(*)$, then $F_d=0$ is the tangent cone at $(0,0)$. Note that this answers immediately and trivially to your first question.
Finally, the method applies to just any point $(x_0,y_0)$ after a simple change of variables $X=x-x_0$, $Y=y-y_0$.