8y^2+4xy+2xz+4yz+8x+6y+4z+5=0 canonical form

Given equation of the surface of 2-order:
$$4 x y + 2 x z + 8 x + 8 y^{2} + 4 y z + 6 y + 4 z + 5 = 0$$
This equation looks like:
$$a_{11} x^{2} + 2 a_{12} x y + 2 a_{13} x z + 2 a_{14} x + a_{22} y^{2} + 2 a_{23} y z + 2 a_{24} y + a_{33} z^{2} + 2 a_{34} z + a_{44} = 0$$
where
$$a_{11} = 0$$
$$a_{12} = 2$$
$$a_{13} = 1$$
$$a_{14} = 4$$
$$a_{22} = 8$$
$$a_{23} = 2$$
$$a_{24} = 3$$
$$a_{33} = 0$$
$$a_{34} = 2$$
$$a_{44} = 5$$
The invariants of the equation when converting coordinates are determinants:
$$I_{1} = a_{11} + a_{22} + a_{33}$$

     |a11  a12|   |a22  a23|   |a11  a13|
I2 = |        | + |        | + |        |
     |a12  a22|   |a23  a33|   |a13  a33|

$$I_{3} = \left|\begin{matrix}a_{11} & a_{12} & a_{13}\\a_{12} & a_{22} & a_{23}\\a_{13} & a_{23} & a_{33}\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}a_{11} & a_{12} & a_{13} & a_{14}\\a_{12} & a_{22} & a_{23} & a_{24}\\a_{13} & a_{23} & a_{33} & a_{34}\\a_{14} & a_{24} & a_{34} & a_{44}\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}a_{11} - \lambda & a_{12} & a_{13}\\a_{12} & a_{22} - \lambda & a_{23}\\a_{13} & a_{23} & a_{33} - \lambda\end{matrix}\right|$$

     |a11  a14|   |a22  a24|   |a33  a34|
K2 = |        | + |        | + |        |
     |a14  a44|   |a24  a44|   |a34  a44|

     |a11  a12  a14|   |a22  a23  a24|   |a11  a13  a14|
     |             |   |             |   |             |
K3 = |a12  a22  a24| + |a23  a33  a34| + |a13  a33  a34|
     |             |   |             |   |             |
     |a14  a24  a44|   |a24  a34  a44|   |a14  a34  a44|

substitute coefficients
$$I_{1} = 8$$

     |0  2|   |8  2|   |0  1|
I2 = |    | + |    | + |    |
     |2  8|   |2  0|   |1  0|

$$I_{3} = \left|\begin{matrix}0 & 2 & 1\\2 & 8 & 2\\1 & 2 & 0\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}0 & 2 & 1 & 4\\2 & 8 & 2 & 3\\1 & 2 & 0 & 2\\4 & 3 & 2 & 5\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda & 2 & 1\\2 & 8 - \lambda & 2\\1 & 2 & - \lambda\end{matrix}\right|$$

     |0  4|   |8  3|   |0  2|
K2 = |    | + |    | + |    |
     |4  5|   |3  5|   |2  5|

     |0  2  4|   |8  2  3|   |0  1  4|
     |       |   |       |   |       |
K3 = |2  8  3| + |2  0  2| + |1  0  2|
     |       |   |       |   |       |
     |4  3  5|   |3  2  5|   |4  2  5|

$$I_{1} = 8$$
$$I_{2} = -9$$
$$I_{3} = 0$$
$$I_{4} = 81$$
$$I{\left(\lambda \right)} = - \lambda^{3} + 8 \lambda^{2} + 9 \lambda$$
$$K_{2} = 11$$
$$K_{3} = -117$$
Because
$$I_{3} = 0 \wedge I_{2} \neq 0 \wedge I_{4} \neq 0$$
then by type of surface:
you need to
Make the characteristic equation for the surface:
$$- I_{1} \lambda^{2} + I_{2} \lambda - I_{3} + \lambda^{3} = 0$$
or
$$\lambda^{3} - 8 \lambda^{2} - 9 \lambda = 0$$
$$\lambda_{1} = 9$$
$$\lambda_{2} = -1$$
$$\lambda_{3} = 0$$
then the canonical form of the equation will be
$$\tilde z 2 \sqrt{\frac{\left(-1\right) I_{4}}{I_{2}}} + \left(\tilde x^{2} \lambda_{1} + \tilde y^{2} \lambda_{2}\right) = 0$$
and
$$- \tilde z 2 \sqrt{\frac{\left(-1\right) I_{4}}{I_{2}}} + \left(\tilde x^{2} \lambda_{1} + \tilde y^{2} \lambda_{2}\right) = 0$$
$$9 \tilde x^{2} - \tilde y^{2} + 6 \tilde z = 0$$
and
$$9 \tilde x^{2} - \tilde y^{2} - 6 \tilde z = 0$$

        2           2                 
\tilde x    \tilde y                  
--------- - --------- + 2*\tilde z = 0
   1/3          3                     

and

        2           2                 
\tilde x    \tilde y                  
--------- - --------- - 2*\tilde z = 0
   1/3          3                     

this equation is fora type hyperbolic paraboloid
- reduced to canonical form