Let $C$ be the conic given by the equation $F(x,y)=ax^2+bxy+cy^2+dx+ey+f=0$. Show that if
$$\begin{vmatrix}
2a&b &d \\
b&2c &e \\
d&e &2f
\end{vmatrix}\neq 0,$$
then $C$ has no singular points.
So I want to show that there are no points $(x,y)$ such that $F(x,y)=\frac{\partial F}{\partial x}(x,y)+\frac{\partial F}{\partial y}(x,y)=0$. I've come at this a couple different ways. First, since the determinant is non zero then this matrix is bijective and thus since its first two rows are precisely the coefficients of $\frac{\partial }{\partial x}$ and $\frac{\partial }{\partial y}$, respectively, this means that I should look at vectors $[x,y,z]$ which it maps to the vectors $[0,0,\lambda]$. This means that the singular point (there can be only one since otherwise the determinant would be zero) must lie on some complex plane through the origin (in $\mathbb{C}^3$ I guess), which maps to the the $x=y=0$ subspace of $\mathbb{C}^3$. This doesn't really seem to go anywhere though, so @#%! it, I'll just set the two partial derivatives equal to zero, solve for $x$ and $y$, and then plug those into $F(x,y)$ and hope I can show it can't be equal to zero while keeping the determinant non-zero. This was a ton of computation but I did it and ended up with a somewhat messy expression in $a,…,e$, but I can't find enough commonality among the terms to say anything useful about its structure.
Thoughts: There must be some reason for their choice of the third row of this matrix, but I'm not sure what it is. It does make the matrix hermitian, and in a way writing this matrix is based on a homogenization of the equations for the partial derivatives, so maybe I'm suppose to look at things in the projective plane?
So this is where I am, can anyone help me? Thanks.
Best Answer
a) The determinant condition you wrote is a necessary and sufficient condition for the corresponding projective conic $\bar C\subset \mathbb P^2(\mathbb C)$ to be non-singular.
The equation of that projective conic is obtained by homogeneizing $F$ and is $$\bar F(x,y,z)=ax^2+bxy+cy^2+dxz+eyz+fz^2=0$$
b) Indeed, in general consider $\Gamma \subset \mathbb P^2(\mathbb C)$ a curve of degree $d$ in the projective plane given by $G$=0 for some homogeneous polynomial $G(x,y,z)\in \mathbb C[x,y,z]$.
A point $P=[a:b:c]\in \mathbb P^2(\mathbb C)$ will be a singular point of $\Gamma$ if and only $$\frac {\partial G} {\partial x} (P)=\frac {\partial G} {\partial y} (P)=\frac {\partial G} {\partial z} (P)=0 \quad (\bigstar)$$ Note that if $P$ satisfies these equations it will automatically be on $\Gamma$ because of Euler's identity for homogeneous polynomials of degree $d$ : $$x\frac {\partial G} {\partial x} (x,y,z)+y\frac {\partial G} {\partial y} (x,y,z)+z\frac {\partial G} {\partial z} (x,y,z)=d\cdot G(x,y,z)$$
c) In your particular question the condition $$\begin{vmatrix} 2a&b &d \\ b&2c &e \\ d&e &2f \end{vmatrix}= 0$$ is a necessary and sufficient for the system $(\bigstar)$ to have a non-zero solution and thus for the existence of a singular point $P\in \bar C$ .
d) Finally note carefully that the vanishing of the determinant does not imply the existence of a singular point in the affine part $\mathbb C^2\subset \mathbb P^2(\mathbb C)$ (the part your question is about) : the conic $x^2-x=0$ is perfectly non-singular in $\mathbb C^2$, although the corresponding determinant vanishes.
That vanishing reflects the singularity of the conic at its only point at infinity $[0:1:0]\in \bar C\setminus C$.