9x2+12xy+4y2-24x-16y+3=0 canonical form

Given equation of the surface of 2-order:
$$12 x y - 24 x + 9 x_{2} - 16 y + 4 y_{2} + 3 = 0$$
This equation looks like:
$$a_{11} x^{2} + 2 a_{12} x y + 2 a_{13} x y_{2} + 2 a_{14} x + a_{22} y^{2} + 2 a_{23} y y_{2} + 2 a_{24} y + a_{33} y_{2}^{2} + 2 a_{34} y_{2} + a_{44} = 0$$
where
$$a_{11} = 0$$
$$a_{12} = 6$$
$$a_{13} = 0$$
$$a_{14} = -12$$
$$a_{22} = 0$$
$$a_{23} = 0$$
$$a_{24} = -8$$
$$a_{33} = 0$$
$$a_{34} = 2$$
$$a_{44} = 9 x_{2} + 3$$
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
$$I_{1} = 0$$

     |0  6|   |0  0|   |0  0|
I2 = |    | + |    | + |    |
     |6  0|   |0  0|   |0  0|

$$I_{3} = \left|\begin{matrix}0 & 6 & 0\\6 & 0 & 0\\0 & 0 & 0\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}0 & 6 & 0 & -12\\6 & 0 & 0 & -8\\0 & 0 & 0 & 2\\-12 & -8 & 2 & 9 x_{2} + 3\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda & 6 & 0\\6 & - \lambda & 0\\0 & 0 & - \lambda\end{matrix}\right|$$

     | 0     -12   |   |0      -8   |   |0     2    |
K2 = |             | + |            | + |           |
     |-12  3 + 9*x2|   |-8  3 + 9*x2|   |2  3 + 9*x2|

     | 0   6     -12   |   |0   0     -8   |   | 0   0    -12   |
     |                 |   |               |   |                |
K3 = | 6   0      -8   | + |0   0     2    | + | 0   0     2    |
     |                 |   |               |   |                |
     |-12  -8  3 + 9*x2|   |-8  2  3 + 9*x2|   |-12  2  3 + 9*x2|

$$I_{1} = 0$$
$$I_{2} = -36$$
$$I_{3} = 0$$
$$I_{4} = 144$$
$$I{\left(\lambda \right)} = - \lambda^{3} + 36 \lambda$$
$$K_{2} = -212$$
$$K_{3} = 1044 - 324 x_{2}$$
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
$$\lambda^{3} - 36 \lambda = 0$$
$$\lambda_{1} = -6$$
$$\lambda_{2} = 6$$
$$\lambda_{3} = 0$$
then the canonical form of the equation will be
$$\tilde y2 \cdot 2 \sqrt{\frac{\left(-1\right) I_{4}}{I_{2}}} + \left(\tilde x^{2} \lambda_{1} + \tilde y^{2} \lambda_{2}\right) = 0$$
and
$$- \tilde y2 \cdot 2 \sqrt{\frac{\left(-1\right) I_{4}}{I_{2}}} + \left(\tilde x^{2} \lambda_{1} + \tilde y^{2} \lambda_{2}\right) = 0$$
$$- 6 \tilde x^{2} + 6 \tilde y^{2} + 4 \tilde y2 = 0$$
and
$$- 6 \tilde x^{2} + 6 \tilde y^{2} - 4 \tilde y2 = 0$$

        2           2                  
\tilde x    \tilde y                   
--------- - --------- - 2*\tilde y2 = 0
   1/3         1/3                     

and

        2           2                  
\tilde x    \tilde y                   
--------- - --------- + 2*\tilde y2 = 0
   1/3         1/3                     

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