x^2+5y^2+z^2+2xy+6xz+2yz+6sqrt(6)x+12sqrt(12)y+6sqrt(6)z=0 forma canónica

Se da la ecuación de superficie de 2 grado:
$$x^{2} + 2 x y + 6 x z + 6 \sqrt{6} x + 5 y^{2} + 2 y z + 24 \sqrt{3} y + z^{2} + 6 \sqrt{6} z = 0$$
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
$$a_{11} = 1$$
$$a_{12} = 1$$
$$a_{13} = 3$$
$$a_{14} = 3 \sqrt{6}$$
$$a_{22} = 5$$
$$a_{23} = 1$$
$$a_{24} = 12 \sqrt{3}$$
$$a_{33} = 1$$
$$a_{34} = 3 \sqrt{6}$$
$$a_{44} = 0$$
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
$$I_{1} = 7$$

     |1  1|   |5  1|   |1  3|
I2 = |    | + |    | + |    |
     |1  5|   |1  1|   |3  1|

$$I_{3} = \left|\begin{matrix}1 & 1 & 3\\1 & 5 & 1\\3 & 1 & 1\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}1 & 1 & 3 & 3 \sqrt{6}\\1 & 5 & 1 & 12 \sqrt{3}\\3 & 1 & 1 & 3 \sqrt{6}\\3 \sqrt{6} & 12 \sqrt{3} & 3 \sqrt{6} & 0\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}1 - \lambda & 1 & 3\\1 & 5 - \lambda & 1\\3 & 1 & 1 - \lambda\end{matrix}\right|$$

     |             ___|   |               ___|   |             ___|
     |   1     3*\/ 6 |   |   5      12*\/ 3 |   |   1     3*\/ 6 |
K2 = |                | + |                  | + |                |
     |    ___         |   |     ___          |   |    ___         |
     |3*\/ 6      0   |   |12*\/ 3      0    |   |3*\/ 6      0   |

     |                       ___ |   |                        ___|   |                      ___|
     |   1        1      3*\/ 6  |   |   5         1     12*\/ 3 |   |   1        3     3*\/ 6 |
     |                           |   |                           |   |                         |
     |                        ___|   |                       ___ |   |                      ___|
K3 = |   1        5      12*\/ 3 | + |   1         1     3*\/ 6  | + |   3        1     3*\/ 6 |
     |                           |   |                           |   |                         |
     |    ___       ___          |   |     ___      ___          |   |    ___      ___         |
     |3*\/ 6   12*\/ 3      0    |   |12*\/ 3   3*\/ 6      0    |   |3*\/ 6   3*\/ 6      0   |
           

$$I_{1} = 7$$
$$I_{2} = 0$$
$$I_{3} = -36$$
$$I_{4} = 4536 - 864 \sqrt{2}$$
$$I{\left(\lambda \right)} = - \lambda^{3} + 7 \lambda^{2} - 36$$
$$K_{2} = -540$$
$$K_{3} = -1188 + 432 \sqrt{2}$$
Como

I3 != 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
$$\lambda^{3} - 7 \lambda^{2} + 36 = 0$$
$$\lambda_{1} = 6$$
$$\lambda_{2} = 3$$
$$\lambda_{3} = -2$$
entonces la forma canónica de la ecuación será
$$\left(\tilde z^{2} \lambda_{3} + \left(\tilde x^{2} \lambda_{1} + \tilde y^{2} \lambda_{2}\right)\right) + \frac{I_{4}}{I_{3}} = 0$$
$$6 \tilde x^{2} + 3 \tilde y^{2} - 2 \tilde z^{2} - 126 + 24 \sqrt{2} = 0$$
$$- \frac{\tilde z^{2}}{\left(\frac{\frac{1}{2} \sqrt{2}}{\frac{1}{\sqrt{126 - 24 \sqrt{2}}}}\right)^{2}} + \left(\frac{\tilde x^{2}}{\left(\frac{\frac{1}{6} \sqrt{6}}{\frac{1}{\sqrt{126 - 24 \sqrt{2}}}}\right)^{2}} + \frac{\tilde y^{2}}{\left(\frac{\frac{1}{3} \sqrt{3}}{\frac{1}{\sqrt{126 - 24 \sqrt{2}}}}\right)^{2}}\right) = 1$$
es la ecuación para el tipo hiperboloide unilateral
- está reducida a la forma canónica