z^2-2sqrt(2)xz-2sqrt(6)x-2sqrt(3)z-1=0 forma canónica

Se da la ecuación de superficie de 2 grado:
$$- 2 \sqrt{2} x z - 2 \sqrt{6} x + z^{2} - 2 \sqrt{3} z - 1 = 0$$
Esta ecuación tiene la forma:
$$a_{11} x^{2} + 2 a_{12} x y + 2 a_{13} x z + 2 a_{14} x + a_{22} y^{2} + 2 a_{23} y z + 2 a_{24} y + a_{33} z^{2} + 2 a_{34} z + a_{44} = 0$$
donde
$$a_{11} = 0$$
$$a_{12} = 0$$
$$a_{13} = - \sqrt{2}$$
$$a_{14} = - \sqrt{6}$$
$$a_{22} = 0$$
$$a_{23} = 0$$
$$a_{24} = 0$$
$$a_{33} = 1$$
$$a_{34} = - \sqrt{3}$$
$$a_{44} = -1$$
Las invariantes de esta ecuación al transformar las coordenadas son los determinantes:
$$I_{1} = a_{11} + a_{22} + a_{33}$$

     |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|

sustituimos coeficientes
$$I_{1} = 1$$

                       |           ___|
     |0  0|   |0  0|   |  0     -\/ 2 |
I2 = |    | + |    | + |              |
     |0  0|   |0  1|   |   ___        |
                       |-\/ 2     1   |

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

     |           ___|             |           ___|
     |  0     -\/ 6 |   |0  0 |   |  1     -\/ 3 |
K2 = |              | + |     | + |              |
     |   ___        |   |0  -1|   |   ___        |
     |-\/ 6     -1  |             |-\/ 3     -1  |

                                                 |           ___     ___|
     |              ___|   |0    0       0   |   |  0     -\/ 2   -\/ 6 |
     |  0     0  -\/ 6 |   |                 |   |                      |
     |                 |   |              ___|   |   ___             ___|
K3 = |  0     0    0   | + |0    1     -\/ 3 | + |-\/ 2     1     -\/ 3 |
     |                 |   |                 |   |                      |
     |   ___           |   |      ___        |   |   ___     ___        |
     |-\/ 6   0    -1  |   |0  -\/ 3     -1  |   |-\/ 6   -\/ 3     -1  |
                                                 

$$I_{1} = 1$$
$$I_{2} = -2$$
$$I_{3} = 0$$
$$I_{4} = 0$$
$$I{\left(\lambda \right)} = - \lambda^{3} + \lambda^{2} + 2 \lambda$$
$$K_{2} = -10$$
$$K_{3} = -16$$
Como
$$I_{3} = 0 \wedge I_{4} = 0 \wedge I_{2} \neq 0$$
entonces por razón de tipos de rectas:
hay que
Formulamos la ecuación característica para nuestra superficie:
$$- I_{1} \lambda^{2} + I_{2} \lambda - I_{3} + \lambda^{3} = 0$$
o
$$\lambda^{3} - \lambda^{2} - 2 \lambda = 0$$
$$\lambda_{1} = 2$$
$$\lambda_{2} = -1$$
$$\lambda_{3} = 0$$
entonces la forma canónica de la ecuación será
$$\left(\tilde x^{2} \lambda_{1} + \tilde y^{2} \lambda_{2}\right) + \frac{K_{3}}{I_{2}} = 0$$
$$2 \tilde x^{2} - \tilde y^{2} + 8 = 0$$
$$\frac{\tilde x^{2}}{4} - \frac{\tilde y^{2}}{8} = -1$$
es la ecuación para el tipo cilindro hiperbólico
- está reducida a la forma canónica