3*y^2+4z^2+sqrt(6)x+2*sqrt(3)*y-2*sqrt(2)z+13=0 forma canónica

Se da la ecuación de superficie de 2 grado:
$$\sqrt{6} x + 3 y^{2} + 2 \sqrt{3} y + 4 z^{2} - 2 \sqrt{2} z + 13 = 0$$
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
$$a_{11} = 0$$
$$a_{12} = 0$$
$$a_{13} = 0$$
$$a_{14} = \frac{\sqrt{6}}{2}$$
$$a_{22} = 3$$
$$a_{23} = 0$$
$$a_{24} = \sqrt{3}$$
$$a_{33} = 4$$
$$a_{34} = - \sqrt{2}$$
$$a_{44} = 13$$
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
$$I_{1} = 7$$

     |0  0|   |3  0|   |0  0|
I2 = |    | + |    | + |    |
     |0  3|   |0  4|   |0  4|

$$I_{3} = \left|\begin{matrix}0 & 0 & 0\\0 & 3 & 0\\0 & 0 & 4\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}0 & 0 & 0 & \frac{\sqrt{6}}{2}\\0 & 3 & 0 & \sqrt{3}\\0 & 0 & 4 & - \sqrt{2}\\\frac{\sqrt{6}}{2} & \sqrt{3} & - \sqrt{2} & 13\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda & 0 & 0\\0 & 3 - \lambda & 0\\0 & 0 & 4 - \lambda\end{matrix}\right|$$

     |         ___|                                    
     |       \/ 6 |                                    
     |  0    -----|   |         ___|   |           ___|
     |         2  |   |  3    \/ 3 |   |  4     -\/ 2 |
K2 = |            | + |            | + |              |
     |  ___       |   |  ___       |   |   ___        |
     |\/ 6        |   |\/ 3    13  |   |-\/ 2     13  |
     |-----   13  |                                    
     |  2         |                                    

     |                ___|                             |                 ___ |
     |              \/ 6 |                             |               \/ 6  |
     |  0      0    -----|   |                 ___ |   |  0      0     ----- |
     |                2  |   |  3      0     \/ 3  |   |                 2   |
     |                   |   |                     |   |                     |
     |                ___|   |                  ___|   |                  ___|
K3 = |  0      3    \/ 3 | + |  0      4     -\/ 2 | + |  0      4     -\/ 2 |
     |                   |   |                     |   |                     |
     |  ___              |   |  ___     ___        |   |  ___                |
     |\/ 6     ___       |   |\/ 3   -\/ 2     13  |   |\/ 6      ___        |
     |-----  \/ 3    13  |      |-----  -\/ 2     13  |
     |  2                |                             |  2                  |
                                  

$$I_{1} = 7$$
$$I_{2} = 12$$
$$I_{3} = 0$$
$$I_{4} = -18$$
$$I{\left(\lambda \right)} = - \lambda^{3} + 7 \lambda^{2} - 12 \lambda$$
$$K_{2} = \frac{169}{2}$$
$$K_{3} = \frac{255}{2}$$
Como
$$I_{3} = 0 \wedge I_{2} \neq 0 \wedge I_{4} \neq 0$$
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
$$\lambda^{3} - 7 \lambda^{2} + 12 \lambda = 0$$
$$\lambda_{1} = 4$$
$$\lambda_{2} = 3$$
$$\lambda_{3} = 0$$
entonces la forma canónica de la ecuación será
$$\tilde z 2 \sqrt{\frac{\left(-1\right) I_{4}}{I_{2}}} + \left(\tilde x^{2} \lambda_{1} + \tilde y^{2} \lambda_{2}\right) = 0$$
y
$$- \tilde z 2 \sqrt{\frac{\left(-1\right) I_{4}}{I_{2}}} + \left(\tilde x^{2} \lambda_{1} + \tilde y^{2} \lambda_{2}\right) = 0$$
$$4 \tilde x^{2} + 3 \tilde y^{2} + \sqrt{6} \tilde z = 0$$
y
$$4 \tilde x^{2} + 3 \tilde y^{2} - \sqrt{6} \tilde z = 0$$
$$2 \tilde z + \left(\frac{\tilde x^{2}}{\frac{1}{8} \sqrt{6}} + \frac{\tilde y^{2}}{\frac{1}{6} \sqrt{6}}\right) = 0$$
y
$$- 2 \tilde z + \left(\frac{\tilde x^{2}}{\frac{1}{8} \sqrt{6}} + \frac{\tilde y^{2}}{\frac{1}{6} \sqrt{6}}\right) = 0$$
es la ecuación para el tipo paraboloide elíptico
- está reducida a la forma canónica