8x^2+7y^2+9z^2-4xy+4xz canonical form

Given equation of the surface of 2-order:
$$8 x^{2} - 4 x y + 4 x z + 7 y^{2} + 9 z^{2} = 0$$
This equation looks like:
$$a_{11} x^{2} + 2 a_{12} x y + 2 a_{13} x z + a_{22} y^{2} + 2 a_{23} y z + a_{33} z^{2} + 2 a_{14} x + 2 a_{24} y + 2 a_{34} z + a_{44} = 0$$
where
$$a_{11} = 8$$
$$a_{12} = -2$$
$$a_{13} = 2$$
$$a_{14} = 0$$
$$a_{22} = 7$$
$$a_{23} = 0$$
$$a_{24} = 0$$
$$a_{33} = 9$$
$$a_{34} = 0$$
$$a_{44} = 0$$
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
$$I_{1} = 24$$

     |8   -2|   |7  0|   |8  2|
I2 = |      | + |    | + |    |
     |-2  7 |   |0  9|   |2  9|

$$I_{3} = \left|\begin{matrix}8 & -2 & 2\\-2 & 7 & 0\\2 & 0 & 9\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}8 & -2 & 2 & 0\\-2 & 7 & 0 & 0\\2 & 0 & 9 & 0\\0 & 0 & 0 & 0\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda + 8 & -2 & 2\\-2 & - \lambda + 7 & 0\\2 & 0 & - \lambda + 9\end{matrix}\right|$$

     |8  0|   |7  0|   |9  0|
K2 = |    | + |    | + |    |
     |0  0|   |0  0|   |0  0|

     |8   -2  0|   |7  0  0|   |8  2  0|
     |         |   |       |   |       |
K3 = |-2  7   0| + |0  9  0| + |2  9  0|
     |         |   |       |   |       |
     |0   0   0|   |0  0  0|   |0  0  0|

$$I_{1} = 24$$
$$I_{2} = 183$$
$$I_{3} = 440$$
$$I_{4} = 0$$
$$I{\left(\lambda \right)} = - \lambda^{3} + 24 \lambda^{2} - 183 \lambda + 440$$
$$K_{2} = 0$$
$$K_{3} = 0$$
Because

I3 != 0

then by type of surface:
you need to
Make the characteristic equation for the surface:
$$- I_{1} \lambda^{2} + \lambda^{3} + I_{2} \lambda - I_{3} = 0$$
or
$$\lambda^{3} - 24 \lambda^{2} + 183 \lambda - 440 = 0$$
Solve this equation
$$\lambda_{1} = 11$$
$$\lambda_{2} = 8$$
$$\lambda_{3} = 5$$
then the canonical form of the equation will be
$$\left(\tilde z^{2} \lambda_{3} + \left(\tilde x^{2} \lambda_{1} + \tilde y^{2} \lambda_{2}\right)\right) + \frac{I_{4}}{I_{3}} = 0$$
$$11 \tilde x^{2} + 8 \tilde y^{2} + 5 \tilde z^{2} = 0$$
$$\frac{\tilde z^{2}}{\left(\frac{\sqrt{5}}{5}\right)^{2}} + \left(\frac{\tilde x^{2}}{\left(\frac{\sqrt{11}}{11}\right)^{2}} + \frac{\tilde y^{2}}{\left(\frac{\sqrt{2}}{4}\right)^{2}}\right) = 0$$
this equation is fora type imaginary cone
- reduced to canonical form