Is a general (non-homogeneous) quadratic equation in $\mathbb{R}^3$ an ellipsoid? This sounds like a simple problem, but I can't get it done. Given the general equation
$ax^2 + by^2 + cz^2 + dxy + exz + fyz + gx + hy + iz + j = 0,$ 
what are the requirements on the coefficients so that this equation represents an ellipsoid? All coefficients are real.
Some more context. To make the representation unique, we establish another constraint: $a + c = 1$. I'm not sure how important that might be. The question is in homework where we have to fit this equation to a set of points, using QR-decomposition. That isn't a problem, the small followup question "how can you see the coefficients determine an ellipsoid" is the problem.
I tried working on the equation to get something usable, from basic term regrouping to putting stuff in matrices and decomposing these, which didn't really help. One theoretic way to do it, I think, is to find the center of mass of the surface, translate to the origin, then find the principal axes, transform again, and show the radii are all larger than 0. Sounds like a lot of work for a simple question, and I'm kind of stuck on the "center of mass" part, I don't feel the idea here is to integrate the formula over $\mathbb{R}^3$. I've also been thinking about calculating curvature or something with the Frenet-Serret formulas, but my geometry courses have been too long ago to know what I want here.
Is there a simple test to determine this? Something along the lines of "the eigenvalues of matrix foo are all positive" or "the value of bar is real" or I don't know what...
Thanks a lot for any insights you can give me!
 A: There are two steps involved. 
Step 1: translate to centre of mass
The goal of translation to the centre of mass is to remove the linear terms in your coefficients. You want to turn your expression into the form
$$ AX^2 + BY^2 + CZ^2 + D XY + E YZ + F XZ + G = 0$$
where $X = (x - x_0)$, $Y = (y - y_0)$ and $Z = (z-z_0)$, where $x_0, y_0, z_0$ are the coordinates for the centre of mass. To do so you can solve forward by completing the square, or solve backwards by plugging in the expressions for $X,Y,Z$ and matching coefficients. 
Step 2: check for positivity of eigenvalues
Relative to the centre of mass, you can write your equation as
$$ Q(v) = -G $$
where $Q$ is the quadratic form 
$$ Q(v) = AX^2 + BY^2 + \cdots + F XZ, \qquad v = (X,Y,Z) $$
Then recall that the level surfaces of a quadratic form define ellipsoids if and only if it is positive definite. So computing the eigenvalues (among the many ways to check that a quadratic form is positive definite) will get you the answer. (Of course, this will also require $G$ to be a negative number for there to be a solution.)
