Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
lol 'cat goes in microwave' – Ethan Jan 20 '13 at 4:15
+1: This is a very interesting, carefully thought out question. (I have removed the tag "elliptic-curves" which has nothing to do with the question : confusingly, an ellipse is not an elliptic curve !) – Georges Elencwajg Jan 20 '13 at 9:34
up vote 3 down vote accepted

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 $[0:1:0]\in \bar C\setminus C$.

share|cite|improve this answer
After pondering this answer all day, and enriching my understanding of the projective plane considerably in the process, I have fully understood the particulars of how affine and projective singularities relate. Thank you. – esproff Jan 21 '13 at 1:32
Dear cat, you are welcome. – Georges Elencwajg Jan 21 '13 at 7:01

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.