3x^2+3y^2-3z^2-6*(2^0.5)x+2*(3^0.5)y+2*(6^0.5)z=-8 forma canónica

Se da la ecuación de superficie de 2 grado:
$$3 x^{2} - 6 \sqrt{2} x + 3 y^{2} + 2 \sqrt{3} y - 3 z^{2} + 2 \sqrt{6} z + 8 = 0$$
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
$$a_{11} = 3$$
$$a_{12} = 0$$
$$a_{13} = 0$$
$$a_{14} = - 3 \sqrt{2}$$
$$a_{22} = 3$$
$$a_{23} = 0$$
$$a_{24} = \sqrt{3}$$
$$a_{33} = -3$$
$$a_{34} = \sqrt{6}$$
$$a_{44} = 8$$
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
$$I_{1} = 3$$

     |3  0|   |3  0 |   |3  0 |
I2 = |    | + |     | + |     |
     |0  3|   |0  -3|   |0  -3|

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

     |               ___|   |         ___|   |         ___|
     |   3      -3*\/ 2 |   |  3    \/ 3 |   | -3    \/ 6 |
K2 = |                  | + |            | + |            |
     |     ___          |   |  ___       |   |  ___       |
     |-3*\/ 2      8    |   |\/ 3     8  |   |\/ 6     8  |

     |                      ___|   |                ___|   |                      ___|
     |   3        0    -3*\/ 2 |   |  3      0    \/ 3 |   |   3        0    -3*\/ 2 |
     |                         |   |                   |   |                         |
     |                    ___  |   |                ___|   |                    ___  |
K3 = |   0        3     \/ 3   | + |  0     -3    \/ 6 | + |   0       -3     \/ 6   |
     |                         |   |                   |   |                         |
     |     ___    ___          |   |  ___    ___       |   |     ___    ___          |
     |-3*\/ 2   \/ 3      8    |   |\/ 3   \/ 6     8  |   |-3*\/ 2   \/ 6      8    |
           

$$I_{1} = 3$$
$$I_{2} = -9$$
$$I_{3} = -27$$
$$I_{4} = -81$$
$$I{\left(\lambda \right)} = - \lambda^{3} + 3 \lambda^{2} + 9 \lambda - 27$$
$$K_{2} = -3$$
$$K_{3} = -108$$
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
$$\lambda^{3} - 3 \lambda^{2} - 9 \lambda + 27 = 0$$
$$\lambda_{1} = -3$$
$$\lambda_{2} = 3$$
$$\lambda_{3} = 3$$
entonces la forma canónica de la ecuación será
$$\left(\tilde z^{2} \lambda_{3} + \left(\tilde x^{2} \lambda_{1} + \tilde y^{2} \lambda_{2}\right)\right) + \frac{I_{4}}{I_{3}} = 0$$
$$- 3 \tilde x^{2} + 3 \tilde y^{2} + 3 \tilde z^{2} + 3 = 0$$
$$- \frac{\tilde x^{2}}{\left(\frac{\frac{1}{3} \sqrt{3}}{\frac{1}{3} \sqrt{3}}\right)^{2}} + \left(\frac{\tilde y^{2}}{\left(\frac{\frac{1}{3} \sqrt{3}}{\frac{1}{3} \sqrt{3}}\right)^{2}} + \frac{\tilde z^{2}}{\left(\frac{\frac{1}{3} \sqrt{3}}{\frac{1}{3} \sqrt{3}}\right)^{2}}\right) = -1$$
es la ecuación para el tipo hiperboloide bilateral
- está reducida a la forma canónica