4x^2+9y^2+36z^2-8x-36y+72z+40=0 canonical form

Given equation of the surface of 2-order:
$$4 x^{2} - 8 x + 9 y^{2} - 36 y + 36 z^{2} + 72 z + 40 = 0$$
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
$$a_{11} = 4$$
$$a_{12} = 0$$
$$a_{13} = 0$$
$$a_{14} = -4$$
$$a_{22} = 9$$
$$a_{23} = 0$$
$$a_{24} = -18$$
$$a_{33} = 36$$
$$a_{34} = 36$$
$$a_{44} = 40$$
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
$$I_{1} = 49$$

     |4  0|   |9  0 |   |4  0 |
I2 = |    | + |     | + |     |
     |0  9|   |0  36|   |0  36|

$$I_{3} = \left|\begin{matrix}4 & 0 & 0\\0 & 9 & 0\\0 & 0 & 36\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}4 & 0 & 0 & -4\\0 & 9 & 0 & -18\\0 & 0 & 36 & 36\\-4 & -18 & 36 & 40\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}4 - \lambda & 0 & 0\\0 & 9 - \lambda & 0\\0 & 0 & 36 - \lambda\end{matrix}\right|$$

     |4   -4|   | 9   -18|   |36  36|
K2 = |      | + |        | + |      |
     |-4  40|   |-18  40 |   |36  40|

     |4    0   -4 |   | 9   0   -18|   |4   0   -4|
     |            |   |            |   |          |
K3 = |0    9   -18| + | 0   36  36 | + |0   36  36|
     |            |   |            |   |          |
     |-4  -18  40 |   |-18  36  40 |   |-4  36  40|

$$I_{1} = 49$$
$$I_{2} = 504$$
$$I_{3} = 1296$$
$$I_{4} = -46656$$
$$I{\left(\lambda \right)} = - \lambda^{3} + 49 \lambda^{2} - 504 \lambda + 1296$$
$$K_{2} = 324$$
$$K_{3} = -10368$$
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
$$\lambda^{3} - 49 \lambda^{2} + 504 \lambda - 1296 = 0$$
$$\lambda_{1} = 36$$
$$\lambda_{2} = 9$$
$$\lambda_{3} = 4$$
then the canonical form of the equation will be
$$\left(\tilde z^{2} \lambda_{3} + \left(\tilde x^{2} \lambda_{1} + \tilde y^{2} \lambda_{2}\right)\right) + \frac{I_{4}}{I_{3}} = 0$$
$$36 \tilde x^{2} + 9 \tilde y^{2} + 4 \tilde z^{2} - 36 = 0$$

        2           2           2    
\tilde x    \tilde y    \tilde z     
--------- + --------- + --------- = 1
        2           2           2    
 /  1  \     /  1  \     /  1  \     
 |-----|     |-----|     |-----|     
 \6*1/6/     \3*1/6/     \2*1/6/     

this equation is fora type ellipsoid
- reduced to canonical form