6x^2+4xy+6y2=1 canonical form

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

     |6  2|   |0  0|   |6  0|
I2 = |    | + |    | + |    |
     |2  0|   |0  0|   |0  0|

$$I_{3} = \left|\begin{matrix}6 & 2 & 0\\2 & 0 & 0\\0 & 0 & 0\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}6 & 2 & 0 & 0\\2 & 0 & 0 & 0\\0 & 0 & 0 & 3\\0 & 0 & 3 & -1\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}6 - \lambda & 2 & 0\\2 & - \lambda & 0\\0 & 0 & - \lambda\end{matrix}\right|$$

     |6  0 |   |0  0 |   |0  3 |
K2 = |     | + |     | + |     |
     |0  -1|   |0  -1|   |3  -1|

     |6  2  0 |   |0  0  0 |   |6  0  0 |
     |        |   |        |   |        |
K3 = |2  0  0 | + |0  0  3 | + |0  0  3 |
     |        |   |        |   |        |
     |0  0  -1|   |0  3  -1|   |0  3  -1|

$$I_{1} = 6$$
$$I_{2} = -4$$
$$I_{3} = 0$$
$$I_{4} = 36$$
$$I{\left(\lambda \right)} = - \lambda^{3} + 6 \lambda^{2} + 4 \lambda$$
$$K_{2} = -15$$
$$K_{3} = -50$$
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
$$\lambda^{3} - 6 \lambda^{2} - 4 \lambda = 0$$
$$\lambda_{1} = 3 - \sqrt{13}$$
$$\lambda_{2} = 3 + \sqrt{13}$$
$$\lambda_{3} = 0$$
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
$$- \tilde y2 \cdot 2 \sqrt{\frac{\left(-1\right) I_{4}}{I_{2}}} + \left(\tilde x^{2} \lambda_{1} + \tilde y^{2} \lambda_{2}\right) = 0$$
$$\tilde x^{2} \left(3 - \sqrt{13}\right) + \tilde y^{2} \left(3 + \sqrt{13}\right) + 6 \tilde y2 = 0$$
and
$$\tilde x^{2} \left(3 - \sqrt{13}\right) + \tilde y^{2} \left(3 + \sqrt{13}\right) - 6 \tilde y2 = 0$$
$$- 2 \tilde y2 + \left(\frac{\tilde x^{2}}{3 \frac{1}{-3 + \sqrt{13}}} - \frac{\tilde y^{2}}{3 \frac{1}{3 + \sqrt{13}}}\right) = 0$$
and
$$2 \tilde y2 + \left(\frac{\tilde x^{2}}{3 \frac{1}{-3 + \sqrt{13}}} - \frac{\tilde y^{2}}{3 \frac{1}{3 + \sqrt{13}}}\right) = 0$$
this equation is fora type hyperbolic paraboloid
- reduced to canonical form