x^2+4*y^2+8*z^2-16 canonical form

Given equation of the surface of 2-order:
$$x^{2} + 4 y^{2} + 8 z^{2} - 16 = 0$$
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
$$a_{11} = 1$$
$$a_{12} = 0$$
$$a_{13} = 0$$
$$a_{14} = 0$$
$$a_{22} = 4$$
$$a_{23} = 0$$
$$a_{24} = 0$$
$$a_{33} = 8$$
$$a_{34} = 0$$
$$a_{44} = -16$$
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
$$I_{1} = 13$$

     |1  0|   |4  0|   |1  0|
I2 = |    | + |    | + |    |
     |0  4|   |0  8|   |0  8|

$$I_{3} = \left|\begin{matrix}1 & 0 & 0\\0 & 4 & 0\\0 & 0 & 8\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}1 & 0 & 0 & 0\\0 & 4 & 0 & 0\\0 & 0 & 8 & 0\\0 & 0 & 0 & -16\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}1 - \lambda & 0 & 0\\0 & 4 - \lambda & 0\\0 & 0 & 8 - \lambda\end{matrix}\right|$$

     |1   0 |   |4   0 |   |8   0 |
K2 = |      | + |      | + |      |
     |0  -16|   |0  -16|   |0  -16|

     |1  0   0 |   |4  0   0 |   |1  0   0 |
     |         |   |         |   |         |
K3 = |0  4   0 | + |0  8   0 | + |0  8   0 |
     |         |   |         |   |         |
     |0  0  -16|   |0  0  -16|   |0  0  -16|

$$I_{1} = 13$$
$$I_{2} = 44$$
$$I_{3} = 32$$
$$I_{4} = -512$$
$$I{\left(\lambda \right)} = - \lambda^{3} + 13 \lambda^{2} - 44 \lambda + 32$$
$$K_{2} = -208$$
$$K_{3} = -704$$
Because

I3 != 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
$$\lambda^{3} - 13 \lambda^{2} + 44 \lambda - 32 = 0$$
$$\lambda_{1} = 8$$
$$\lambda_{2} = 4$$
$$\lambda_{3} = 1$$
then the canonical form of the equation will be
$$\left(\tilde z^{2} \lambda_{3} + \left(\tilde x^{2} \lambda_{1} + \tilde y^{2} \lambda_{2}\right)\right) + \frac{I_{4}}{I_{3}} = 0$$
$$8 \tilde x^{2} + 4 \tilde y^{2} + \tilde z^{2} - 16 = 0$$

        2            2           2    
\tilde x     \tilde y    \tilde z     
---------- + --------- + --------- = 1
         2           2         2      
//  ___\\     /  1  \      / 1\       
||\/ 2 ||     |-----|      \4 /       
||-----||     \2*1/4/                 
|\  4  /|                             
|-------|                             
\  1/4  /                             

this equation is fora type ellipsoid
- reduced to canonical form