<s>[INST] 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..
 [/INST] 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 
$
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