# Show that if the curve $y^2 = p(x)$ has a double point, then it must be of the form $(r,0)$ where $r$ is a double root of $p(x)$.

Let $p(x) = ax^3 + bx^2 + cx + d$ where $a,b,c,d \in\mathbb{R}$. Show that if the curve $y^2 = p(x)$ has a double point, then it must be of the form $(r,0)$ where $r$ is a double root of $p(x)$.

$$p(r) = p'(r) = 0 \implies (x-r)^2\mid p(x)$$

$$p(x) = y^2 \implies p'(x) = 2y \text{ or } y'$$

I am stuck and don't know where or how to go from here.

-
If you are taking derivatives on both sides of $p(x)=y^2$, then the answer should be $p'(x) = 2yy'$, not "$p'(x)=2y$ or $p'(x)=y'$". – Arturo Magidin Jul 18 '12 at 20:12

A cubic curve cannot have two double points, unless it's the union of a line and a conic, or of three lines. Indeed the line joining the two double points would have (at least) 4 intersections with the cubic.

-

A point $(a,b)$ on $y^2 - ax^3 - bx^2 -cx - d$ is a simple point if and only if $\frac{\partial}{\partial x}(y^2-p(x))\Bigm|_{(x,y)=(a,b)}\neq 0$ or $\frac{\partial}{\partial y}(y^2-p(x))\Bigm|_{(x,y)=(a,b)} \neq 0$. It is a multiple point if and only if both partial derivatives are zero.

Suppose that $(a,b)$ is not a simple point. Then \begin{align*} \frac{\partial}{\partial x}(y^2-p(x))\Bigm|_{(x,y)=(a,b)} = p'(a) &=0\\ \frac{\partial}{\partial y}(y^2-p(x))\Bigm|_{(x,y)=(a,b)} = 2b&=0. \end{align*} From the second equation, we have that $b=0$. Hence, $0^2 = p(a)$, so $a$ is a root of both $p(x)$ and $p'(x)$, hence $a$ is a multiple root of $p(x)$. This proves that if $(a,b)$ is a multiple point, then $b=0$ and $a$ is a multiple root of $p(x)$.

-

Consider the curve $y^2=p(x)$ as a level curve of $f(x,y)=y^2-p(x)$. A double point $(x_0,y_0)$ of the curve must be a singular point for $f$ and so the gradient of $f$ at that point must be zero. Now $\nabla f(x,y)=(-p'(x),2y)$, which implies $p'(x_0)=0$ and $y_0=0$. Since $(x_0,y_0)$ is on the curve, we have $0=y_0^2=p(x_0)$ and so $x_0$ is a double root of $p$, being a root of both $p$ and $p'$.

-