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$\newcommand{\C}{\mathcal{C}}$ 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?

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