Note that if $F$ is homogeneous, since it comes from a bi-linear form, it has quadratic terms and so all terms must be quadratic. So $F$ is of the form
$$
F(X_1, \dots, X_{n+1}) = \sum_{i = 1}^{n+1}a_iX_i^2 + \sum_{i,j}b_{i,j}X_iX_j.
$$
In particular, we have that
$$
\frac{\partial F}{\partial X_k}(X_1,\dots,X_{n+1}) = 2X_k + \sum_jb_{k,j}X_j.
$$
and so $0 = F(0) = \nabla F(0)$. This means that the quadric associated with $F$ has a singular point and so it is affinely equivalent to one of the form
$$
A = \sum_{i = 1}^p X_i^2 - \sum_{j = p+1}^rX_j^2
$$
Let $f : x \in \mathbb{R}^{n+1} \mapsto Dx + b \in \mathbb{R}^{n+1}$ be the affine transformation so that $Af = F$ and in particular, one quadric is mapped to the other. Since $A$ and $F$ are homogenous, they are of the form
$$
A = x^tQx, \quad F = x^tBx
$$
and so since $Af = F$ we have that, as polynomials, their cuadratic terms must coincide. Factoring the composite of $A$ and $f$, then, we get
$$
A = x^tQx = x^tD^tBDx = (Dx)^tBDx = F(Dx)
$$
and thus we can keep the linear part of the isomorphism, which will still behave as desired.
In general the approach $C = (P^{\top}Q^{-1}P)^{-1}$ is valid, however there are some intricacies that need to be dealt with carefully:
- Converting a 3d Gaussian $(\Sigma, \mu)$ into a Quadric ($4\times4$ symmetric matrix $Q$)
- Converting the Conic ($3\times3$ symmetric matrix $C$) back to a Gaussian $(\Sigma', \mu')$
- Sampling the 3D and 2D Gaussians
1. Quadric from Gaussian
I cite my other post that deals with this. To describe the surface-ellipsoid of a 3D Gaussian thresholded at $\sigma$, use this parametrization of $Q$:
$$(x-\mu)^\top\Sigma^{-1}(x-\mu)-\sigma=0$$
$$Q = \begin{bmatrix}A & b \\ b^⊤ & c\end{bmatrix}, \quad
\hat{x}^{\top}Q\hat{x}=0, \quad
\hat{x}=[x^{\top},1]^{\top}$$
$$A=\Sigma^{-1}, \quad b=-A\mu , \quad c=\mu^⊤\Sigma^{-1}\mu-\sigma$$
2. Gaussian from Conic
The conic $C$ is derived from $Q$ by $C = (P^{\top}Q^{-1}P)^{-1}$. To represent is as a 2D Gaussian distribution $(\Sigma', \mu')$ thresholded at $\sigma$, we derive the parameters analogously to above.
$$C = \begin{bmatrix}a&b/2&d/2\\b/2&c&e/2\\d/2&e/2&f\end{bmatrix} = \begin{bmatrix}A'&b'\\b'^{\top} & f\end{bmatrix}$$
$$\Sigma=A^{-1}, \quad \mu =-\Sigma b', \quad \sigma = \mu^{\top} \Sigma^{-1} \mu - f$$
What I have overlooked so far is to scale the Covariance by $\sigma$ (at least thats my interpretation of it, but please correct me if I am wrong):
$$\Sigma'=\Sigma\sigma$$
3. Sampling the $\sigma$-surface/line of a 3D/2D Gaussian
Finding points on the surface of a 3D Gaussian or analogously the circumference of a 2D Gaussian can be done by using the Cholesky Factor of the Covariance. I like to think of it as the square root for matrices:
$$\Sigma=LL^{\top}$$
Points on the Ellipsoid/Ellipse at $\sigma=1$ can now be obtained by transforming points on the unit-sphere/circle $\hat{x}$ like so:
$$x=L\hat{x}+\mu'$$
Sampling directly using $\Sigma$ will also produce an ellipsoid/ellipse, but the radii of the resulting object will be squared and thus not correspond with the formulation of the quadric/conic $Q$/$C$.
Best Answer
In what way did you arrive at that canonical form? It sounds to me as though you had essentially diagonalized the matrix of the quadratic form. You can do that, but doing so corresponds to a projective transformation so it will change the relationship between the affine space and the quadric. Therefore I'd suggest not doing this here.
Instead of merely going for $x\neq 0$ you might as well pick one specific representative and assume $x=1$. So you are looking for points $(y,w,z)\in\mathbb A^3$ with $y+yw-w^2=0$. This is independent of $z$, so you can plot this as a planar curve and then extrude it along the $z$ axis. To graph the planar curve, a hyperbola, you can compute asymptotes and probably a few values, then judge a reasonable interpolation in between. Unless you have learned different techniques.