# Am I right in this proof of a criterion for the nonsingularity of a conic curve?

$$\newcommand{\C}{\mathcal{C}}$$ This is an exercise in Silverman and Tate's Rational Points on Elliptic Curves:

Let $$\C$$ be the conic given by the equation $$F(x,y)=ax^2+bxy+cy^2+dx+ey+f=0.$$ Let $$\delta$$ be the determinant $$\delta:=\begin{vmatrix}2a&b&d\\ b&2c&e\\ d&e&2f\end{vmatrix}.$$ Show that if $$\delta\neq0$$, then $$\C$$ has no singular points; i.e. no $$(x,y)$$ satisfy $$F(x,y)=F_x(x,y)=F_y(x,y)=0$$.

Here, $$F_x$$ and $$F_y$$ are the partial derivatives of $$F$$. I think I've already solved the problem, but am quite unsure of the proof, and hence I would appreciate if somone can check it for me. Here's my attempted proof. Because there's a $$3\times 3$$ whose determinant is being taken, this motivates the introduction of a third variable $$z$$, so I thought to define: $$G(x,y,z):=ax^2+bxy+cy^2+dxz+eyz+fz^2.$$ Note that $$G(x,y,1)\equiv F(x,y)$$ under this definition. Then consider the sequence of equations given by setting each partial derivative of $$G$$ to $$0$$: $$\begin{cases} G_x=2ax+by+dz=0,\\ G_y=bx+2cy+ez=0,\\G_z=dx+ey+2fz=0. \end{cases}$$ Now if $$\mathbf M$$ is the matrix in the question, then $$\delta=\det \mathbf M$$, and we now have an equation of the form $$\mathbf{Mx}=\mathbf 0,$$ which looks right! So we use the condition $$\delta\neq 0$$ we get $$\mathbf x=\mathbf M^{-1}\mathbf 0=\mathbf 0$$, hence $$z=0$$, which is a contradiction, because we need $$z=1$$ for $$(x,y)$$ to be a singular point of $$F(x,y)=G(x,y,1)$$.

I'm not confident of this last part in italics. Can I really just impose the restriction $$z=1$$ at the end? Also, maybe a bigger issue is, how can I just introduce the extra condition $$G_z=0$$, when that's not required (I think) for $$F(x,y)$$ to have singular point $$(x,y)$$? I never used the condition $$F(x,y)=0$$ either, which feels really fishy. Can someone help explaining if my steps were valid, and if not, what the right idea is?