First problem in Silverman's Arithmetic of Elliptic Curves I started working through Silverman's Arithmetic of Elliptic Curves. For some reason it looks like the first problem in the first chapter is the hardest problem in the whole chapter or I'm completely missing something. The Problem statement is the following:

Characterize the values of $A,B$ for which the variety $Y^2Z+AXYZ+BYZ^2=X^3$ is singular over an algebraically closed field.

This should be a fairly trivial problem, but I can't seem to get a "nice" answer. First, I want to look for singular points in the set $Z\neq 0$. Dehomogenizing, we get
$$Y^2+AXY+BY=X^3$$
To compute singular points, we get the following system:
$$\left\{\begin{array}{l}
Y^2+AXY+BY=X^3\\
AY=3X^2\\
2Y+AX+B=0
\end{array} \right.$$
I can solve for $Y$ in the last equation, which gives
$$Y = -\frac{AX+B}{2}$$
Plugging this in to the rest of the system gives
$$\left\{\begin{array}{l}
B^2+2ABX+X^2(A^2+4X)=0\\
6X^2-A(B+AX)=0
\end{array} \right.$$
Now I'm not really sure what to do, since I would need to show that only for specific values of $A,B$ is this system consistent. For example if $B=0$, then $A$ can be anything, since we can choose $X=0$ However, if $B\neq 0$, I'm clueless. I've tried other approaches for eliminating variables, but for each method I seem to get stuck with something ugly.
 A: OK, I actually managed to solve this, so instead of deleting the question, here's the answer. It's easy to see that if $A=0$, then $X=0$, so that in that case, the first equation gives $Y^2+BY=0$ and the third gives $2Y+B=0$. The only solution to this is $Y=0=B$. On the other hand, if $B=0$, then $A$ can be anything and we still have a singular point. So let's assume $B\neq 0$, which implies $A\neq 0$.
From the middle system set $Y=3X^2/A$. Substituting and simplifying to the other two, we get
$$\left\{
\begin{array}{l}
3AB+2A^2X+9X^2=0\\
AB+A^2X+6X^2=0
\end{array}
\right.
$$
Multiply the first equation by $2$ and and subtract $3$ times the second. This gives
$$\left\{
\begin{array}{l}
3AB+A^2X=0\\
AB+A^2X+6X^2=0
\end{array}
\right.
$$
Now subtract the first equation from the second and divide the first equation by the common factor $A$, which gives
$$\left\{
\begin{array}{l}
3B+AX=0\\
-2AB+6X^2=0
\end{array}
\right.
$$
From the first equation, we get $X=-3B/A$. Plugging in to the second equation, we get
$$-2AB+6\frac{9B^2}{A}=0$$
This can be equivalently written as
$$B(A^3-27B)=0$$
This analysis holds assuming that $B\neq 0$. However, $B=0$ as a choice also gave a singular variety and it's included in the solution set to the above equation.
I assume that this is an example of what is called a moduli space? I.e. that the set of parameters that parametrize a property is a variety by itself?
