uxy+uxy-2uxy-3uy-6y=0 canonical form

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

     |0  0|   | 0    -3/2|   |0  0|
I2 = |    | + |          | + |    |
     |0  0|   |-3/2   0  |   |0  0|

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

     |0  0|   |0   -3|   |0  0|
K2 = |    | + |      | + |    |
     |0  0|   |-3  0 |   |0  0|

     |0  0   0 |   | 0    -3/2  -3|   |0  0  0|
     |         |   |              |   |       |
K3 = |0  0   -3| + |-3/2   0    0 | + |0  0  0|
     |         |   |              |   |       |
     |0  -3  0 |   | -3    0    0 |   |0  0  0|

$$I_{1} = 0$$
$$I_{2} = - \frac{9}{4}$$
$$I_{3} = 0$$
$$I_{4} = 0$$
$$I{\left(\lambda \right)} = - \lambda^{3} + \frac{9 \lambda}{4}$$
$$K_{2} = -9$$
$$K_{3} = 0$$
Because
$$I_{3} = 0 \wedge I_{4} = 0 \wedge I_{2} \neq 0$$
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
$$\lambda^{3} - \frac{9 \lambda}{4} = 0$$
$$\lambda_{1} = - \frac{3}{2}$$
$$\lambda_{2} = \frac{3}{2}$$
$$\lambda_{3} = 0$$
then the canonical form of the equation will be
$$\left(\tilde x^{2} \lambda_{1} + \tilde y^{2} \lambda_{2}\right) + \frac{K_{3}}{I_{2}} = 0$$
$$- \frac{3 \tilde x^{2}}{2} + \frac{3 \tilde y^{2}}{2} = 0$$
$$\frac{\tilde x^{2}}{\left(\frac{\sqrt{6}}{3}\right)^{2}} - \frac{\tilde y^{2}}{\left(\frac{\sqrt{6}}{3}\right)^{2}} = 0$$
this equation is fora type two intersecting planes
- reduced to canonical form