Our target is to go on from the general equation of a quadric $$q(x,y,z)=0$$ to the canonical equations of the quadrics. To do this, we will first obtain the socalled limited forms, and from this we will obtain the canonical equations.
Obtaining limited equations
We will say that a quadratic polynomial $$q(x,y,x)=0$$ is reduced, or that the quadric $$Q$$ is reduced, if its form is one of the following ones:
 CENTRED: $$\mu_1x^2+\mu_2y^2+\mu_3z^2+\mu$$, with $$\mu_1,\mu_2>0$$ and $$\mu_3\neq 0$$
 PARABOLIC: $$\mu_1x^2+\mu_2y^22z$$, with $$\mu_1>0$$ and $$\mu_2\neq 0$$
 CYLINDRICAL CENTRED: $$\mu_1 x^2+\mu_2y^2+\mu$$, with $$\mu_1>0$$ and $$\mu_2\neq 0$$
 CYLINDRICAL PARABOLIC: $$\mu_1x^22z$$, with $$\mu_1>0$$
 PARALLEL PLANES: $$\mu_1x^2+\mu$$, with $$\mu_1>0$$
Now our target is to convert a general equation of a conical to one that is of some previous type. To obtain it, the following result is used:
"Given a system of rectangular coordinates $$X=(x,y,z)$$ and a quadratic polynomial $$q(x,y,z)$$, a system of rectangular coordinates $$X'=(x',y',z')$$ exists such that the main part of the polynomial $$q(x',y',z')$$ has the form $$$\lambda_1 x'^2+\lambda_2 y'^2+\lambda_3 z'^3$$$ with $$\lambda_1,\lambda_2$$ and $$\lambda_3 \in \mathbb{R}$$ which we will call a diagonal form.
Also,$$\lambda_1,\lambda_2$$ and $$\lambda_3$$ are the eigenvalues of the main matrix of $$q (x, y, z)$$.
Then, given a general equation of a quadric of the form $$q(x,y,z)=0$$, we calculate first the associated main matrix $$A$$.
As soon as the main matrix associated with the quadric is obtained, we calculate the characteristical polynomial to compute the eigenvalues of the above mentioned matrix.
Let's remember that to calculate the characteristical polynomial, it is necessary to calculate the determinant $$det(A \lambda I)$$.
As soon as the eigenvalues are obtained, from the result announced in advance, we know that a change of variables exists which takes us from the general equation to an equation of the form $$$\lambda_1 x^2+\lambda_2y^2+\lambda_3z^2+2px+2qy+2rz+d$$$
As soon as the first reduction is obtained, we are going to consider different cases for obtaining different limited forms.

If $$\lambda_1\lambda_2\lambda_3\neq 0$$, we obtain an equation of the form $$$\lambda_1 x^2+\lambda_2 y^2+\lambda_3 z^2+2px+2qy+2rz+d$$$ after completing the square three times and, if necessary, a change of sign and rearranging terms, we arrive at a limited form of the centered type.

If there are exactly two $$\lambda_i$$ other than zero, we can suppose that $$\lambda_3=0$$ and $$\lambda_1\lambda_2 \neq 0$$. Again, completing the square, we arrive at a polynomial of the form $$$\lambda_1 x^2+\lambda_2 y^2+2rz+d$$$ Now two new cases arise:
 If $$r = 0$$, it is clear that we obtain, perhaps after rearranging the coordinates and changing signs, a limited form of centered cylindrical type.
 If $$r \neq 0$$, the change of $$z$$ for $$zd/2r$$ leads us to an equation that has the form $$$\lambda_1x^2+\lambda_2y^2+2rz$$$ with $$r \neq 0$$, which obviously is equivalent to a limited form of parabolic type.

Finally, if there is just one nonzero eigenvalue, we can suppose that this proper value is $$\lambda_1$$. Completing the square with regard to this proper value, we can suppose that the equation will have the form $$$\lambda_1x^2+2qy+2rz+d$$$ From here, again two possible cases arise:
 If $$q = r = 0$$, the polynomial is clearly equivalent to the limited form of a pair of parallel planes.
 If one of the two values is other than $$0$$, we can eliminate $$d$$ making the change $$$y \longrightarrow y  \frac{d}{2q}$$$ if $$q \neq 0$$ or the change $$z \longrightarrow z \frac{d}{2r}$$, if $$r \neq 0$$.
Finally, we can change the rectangular coordinates so that $$$\displaystyle \Big(x, \frac{1}{p}(qy+rz), \frac{1}{p}(ry+qz)\Big) \longrightarrow (x',y',z')$$$ where $$$\displaystyle p=(q^2+r^2)^\frac{1}{2}$$$
This leads to a limited form of parabolic cylindrical type.
Therefore, we have managed to take a general equation of a quadric to an equation of the limited type. Next, we will see that from a limited equation, a canonical equation is obtained and hence, it will be possible to deduce what conical it is.
Considering the equation of the quadric $$$q(x,y,z)=x^2+y^2+z^2+2yz+2x+2=0$$$ we are going to think about what its associated limited form is.
To start, let's calculate what its the associated main matrix is: $$$A=\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 1 & 1\end{bmatrix} \Rightarrow det(AxI)=x^3+3x^22x=x(x^23x+2)$$$ $$$=x(x1)(x2)$$$ Therefore, in view of the results, we see that two eigenvalues are not zero and one is zero. The equation of the quadric has become $$$q(x,y,z)=x^2+2y^2+2x+2=0$$$ Since we only have a linear term in $$x$$, completing squares for $$x$$, we see that the equation is of the form $$$q(x,y,z)=(x+1)^2+2y^2+1\approx x^2+2y^2+1=0$$$ Therefore, in view of the equation, we see that it is the limited form of the centered cylindrical type.
Canonical equations
We are going to look for the different canonical equations. This will allow us to classify any type of quadric.
To obtain the different types of canonical equations, we will focus on each of the limited forms and according to the values that the different parameters take, we will obtain the different canonical equations.
Quadrics of centered type
If $$\mu \neq 0$$ and we define positive real numbers $$a, b, c$$ given by $$$\displaystyle a=\sqrt{\frac{\mu}{\mu_1}}, b=\sqrt{\frac{\mu}{\mu_2}}$$$ and $$$\displaystyle c=\sqrt{\frac{\mu}{\mu_3}}$$$ and the limited equation takes the form $$$\displaystyle \frac{x^2}{a^2}+\frac{y^2}{b^2} \pm \frac{z^2}{c^2} \pm 1=0$$$ and the four possibilities for the signs lead us to four quadrics: the election $$(+, +)$$ gives an imaginary ellipsoid; $$(+,)$$ gives a real ellipsoid; $$(, +)$$ gives a non ruled hyperboloid; and $$(,)$$ gives a ruled hyperboloid.
On the other hand, if $$\mu = 0$$, we define the positive real numbers $$a, b, c$$ given by $$$\displaystyle a=\sqrt{\frac{1}{\mu_1}}, b=\sqrt{\frac{1}{\mu_2}}$$$ and $$$c=\sqrt{\frac{1}{\mu_3}}$$$ with which the limited equation is written as: $$$\displaystyle \frac{x^2}{a^2}+\frac{y^2}{b^2} \pm \frac{z^2}{c^2} =0$$$
The election of the positive sign gives us an imaginary cone and the negative sign gives us a real cone.
Quadrics of parabolic type
We define real numbers $$a$$ and $$b$$ given by $$$\displaystyle a=\sqrt{\frac{1}{\mu_1}}$$$ and $$$ \displaystyle b=\sqrt{\frac{1}{\mu_2}}$$$
Then, the limited equation takes the form $$$\displaystyle \frac{x^2}{a^2} \pm \frac{y^2}{b^2} 2z=0$$$
The election of the positive sign gives us an elliptical paraboloid, and the negative sign gives us a hyperbolic paraboloid.
Centered cylinders
Let's first suppose that $$\mu \neq 0$$. In this case, if we define positive real numbers given by $$$\displaystyle a=\sqrt{\frac{\mu}{\mu_1}}$$$ and $$$\displaystyle b=\sqrt{\frac{\mu}{\mu_ 2}}$$$ the limited equation adopts the form $$$\displaystyle \frac{x^2}{a^2} \pm \frac{y^2}{b^2} \pm 1=0 $$$
The election $$(+, +) $$ of the signs gives us an imaginary elliptical cylinder; $$(+,)$$ gives a real elliptical cylinder; $$(, +)$$ and $$(,)$$ give hyperbolic cylinders.
If $$\mu =0$$, we can define positive real numbers $$a$$ and $$b$$ given by $$$\displaystyle a=\sqrt{\frac{1}{\mu_1}}$$$ and $$$\displaystyle b=\sqrt{\frac{1}{\mu_2}}$$$ with which the limited equation takes the form $$$\displaystyle \frac{x^2}{a^2} \pm \frac{y^2}{b^2}=0 $$$
The election of the positive sign gives us a pair of combined imaginary planes and the election of the negative sign gives us a pair of real planes.
Parabolic cylinders
If we set $$\displaystyle p= \frac{1}{\mu_1}$$ , the limited equation is written as $$x^22py=0$$ which is a parabolic cylinder.
Pair of parallel planes
Let's suppose, first, that $$\mu \neq 0$$. If we set $$$ \displaystyle k=\sqrt{\frac{\mu}{\mu_1 }}$$$ the limited equation takes the form $$x^2 \pm k^2 = 0$$, $$k>0$$.
The election of the negative sign gives us a pair of parallel planes, and the positive sign gives us a pair of combined imaginary parallel planes.
If $$\mu=0$$, the limited equation is equivalent to an equation of the form $$x^2=0$$, which represents a double plane (two coincidental parallel planes).
Now, we are going to give a small summary of how we have to proceed to be able to give an affine classification of the quadrics using the previous results.
 To start, considering the general equation of the quadric, we calculate the associate main matrix $$A'$$. As soon as the matrix is obtained, we calculate the associate characteristic polynomial and look for its roots to find the eigenvalues of the matrix $$A'$$. Therefore, the equation of the quadric will be $$\lambda_1x^2+\lambda_2y^2+\lambda_3z^2+2px+2qy+2rz+d$$.
 Then, if some of the eigenvalues are zero, we will complete squares with the coordinates which eigenvalues are nonzero and which have linear terms. This will allow us to obtain one of the limited forms.
 $$\mu_1x^2+\mu_2y^2+\mu_3z^2+\mu$$, with $$\mu_1, \mu_2>0$$, $$\mu_3 \neq 0$$.
 $$\mu_1x^2+\mu_2y^22z$$, with $$\mu_1>0$$, $$\mu_2 \neq 0$$.
 $$\mu_1x^2+\mu_2y^2+\mu$$, with $$\mu_1>0$$, $$\mu_2 \neq 0$$.
 $$\mu_1x^22z$$, with $$\mu_1>0$$.
 $$\mu_1x^2\mu$$, with $$\mu_1>0$$.
 Once we have one of the different limited forms, using the different changes of coordinates exhibited in the section on Canonical Equations, we will obtain one of the canonical equations. Once obtained, we will have converted a general equation into a canonical equation and, therefore, we will have classified the quadric.
Let $$$q(x,y,z)=3x^2+2y^2+z^2+1=0$$$ we are going to classify it.
As is possible to observe, the first reduction will not be necessary thanks to the fact that the main matrix $$A'$$ is already diagonal. Also, since all its eigenvalues are other than zero, the limited form of the quadric is of the centered type.
Finally, from the development of the quadrics of the centered type, we see that three positive real numbers exist, $$a, b, c$$, so that: $$$\displaystyle a=\sqrt{\frac{1}{3}}, b=\sqrt{\frac{1}{2}} \mbox{ and } c=1$$$ Therefore, the canonical equation is $$$\displaystyle \frac{x^2}{a^2}+\frac{y^2}{b ^2}+\frac{z^2}{c^2}=1$$$ with $$\displaystyle a=\sqrt{\frac{1}{3}}, b=\sqrt{\frac{1}{2}}$$ and $$c=1$$. This is an imaginary ellipsoid.
Considering the quadric $$$q(x,y,z)=x^2+4xy+2xz+4y^2+4yz+z^2+4x=0$$$ find its canonical equation so that you can classify it.
First, we calculate its main matrix and the characteristical polynomial associated with the above mentioned matrix: $$$A'=\begin{bmatrix} 1 & 2 & 1 \\ 2 & 4 & 2 \\ 1 & 2 &1 \end{bmatrix} \Longrightarrow det(A'xI)=x^36x^2=0$$$ Then, the roots of the characteristical polynomial are $$\lambda_2=\lambda_3=0$$ and $$\lambda_1=6$$.
Therefore, the equation of the conical is of the form $$$q (x, y, z) =6x^2+4x=0$$$ Since we only have quadratic and linear term in $$x$$, we complete squares for $$x$$. To do so, we change variable $$x'=x\frac{1}{9}$$ , so that the equation of the quadric happens to be of the form $$$q (x, y, z) =3x^2\frac{1}{9}=0$$$ Obviously, in view of the equation, the quadric is a pair of parallel planes.