Show that the equation $x^2+3y^2+4yz-6x+8y+8=0$ becomes a surface. Show that the equation $x^2+3y^2+4yz-6x+8y+8=0$ becomes a surface generated by the movement of a line and explicite its rectilinear generatrices.
I have tried to make its matrix, knowing that $a_{11}=1$, $a_{12}=3$...$a_{00}=8$.. But I don't think that's the way to solve it because the matrix that i will get will have a lot of 0's..
 A: If you want to make a matrix do it like this.
$\mathbf x^T \begin {bmatrix} 
1 &0 &0\\
0 &3 &2\\ 
0 &2 &0\\ \end{bmatrix}\mathbf x + \begin {bmatrix} -6&8&0\end{bmatrix} \mathbf x = 0$
Since that matrix is symmetric is is diagonalizable with ortho-normal basis.
$\mathbf x^t P^T D P \mathbf x + BP^TP\mathbf x = -8\\
\mathbf u = P \mathbf x$
$P = \begin {bmatrix} 
1 &0 &0\\
0 &\frac 2{\sqrt5} &-\frac 1{\sqrt5}\\ 
0 &\frac 1{\sqrt5} &\frac 2{\sqrt5}\\ \end{bmatrix}$
$D = \begin {bmatrix} 
1 &0 &0\\
0 &4 &0\\ 
0 &0 &-1\\ \end{bmatrix}$
$u_1^2 + 4u_2^2 - u_3^2 - 6u_1 + \frac {16}{\sqrt5} u_2 +\frac {8}{\sqrt5} u_3=0\\
(u_1-3)^2 + 4(u_2+\frac2{\sqrt 5})^2 - (u_3- \frac 4{\sqrt 5})^2 = 1 
$
That is a hyperboliod of one sheet.
$(x-3)^2 + 4(\frac 2{\sqrt5} y - \frac 1{\sqrt5} z+\frac2{\sqrt 5})^2 - (\frac 1{\sqrt5} y + \frac 2{\sqrt5}z- \frac 4{\sqrt 5})^2 = 1 
$
A: Here is a solution for the second part (explicitation of a family of generatrices)
Once you have the equation of surface $(S)$ under the form 
$$\tag{1} (S) \  \ \ X^2+4Y^2-Z^2=0 \ \ \ \ \iff \ \ \ \ 4Y^2=Z^2-X^2 $$
(see solution by @Doug M), here is how one finds a system of generatrices. 
Let us consider:
$$\tag{2}(G_{\lambda}) \ \begin{cases}2Y=\lambda(Z-X)\\2Y=\dfrac{1}{\lambda}(Z+X)\end{cases} \ \ \ \ \lambda \in \mathbb{R^*}$$
For each fixed value of $\lambda$, $(G_{\lambda})$ represents a line (Recall that in $\mathbb{R^3}$, a line has two equations, i.e., is represented as the intersection of 2 planes). Moreover :
$$\tag{3} (x,y) \ \text{verifies} \  (2) \ \implies \  (x,y)  \ \text{verifies} \  (1)$$
(by equating the product of the LHSides and RHsides). What is the meaning of $(3)$ ? Plainly that 
$$(G_{\lambda}) \subset (S).$$
Conversely, it is easy to prove that, if $(x,y) \in (S)$, there exist a unique $\lambda$ such that $(x,y) \in (G_{\lambda}).$
We have thus proved that $(S)$ is a ruled surface, "ruled" by generatrices $(G_{\lambda})$.
But that's not all the story; there is a second system of generatrices. 
Let us permute in (2) the RHS, this time with a different parameter, $\mu$.
We get the family of lines:
$$(\Gamma_{\mu}) \ \begin{cases}2Y=\mu(Z+X)\\2Y=\dfrac{1}{\mu}(Z-X)\end{cases}  \ \ \ \ \mu \in \mathbb{R^*}$$
which can as well define surface $(S)$ as a ruled surface.
This double way for the ruling of an hyperboloid with one sheet is nicely represented as an animation in (https://en.wikipedia.org/wiki/Hyperboloid)
