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

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

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

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

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

$$I_{1} = 7$$
$$I_{2} = 12$$
$$I_{3} = 0$$
$$I_{4} = -162$$
$$I{\left(\lambda \right)} = - \lambda^{3} + 7 \lambda^{2} - 12 \lambda$$
$$K_{2} = \frac{149}{2}$$
$$K_{3} = \frac{99}{2}$$
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
$$- \tilde z 2 \sqrt{\frac{\left(-1\right) I_{4}}{I_{2}}} + \left(\tilde x^{2} \lambda_{1} + \tilde y^{2} \lambda_{2}\right) = 0$$
$$4 \tilde x^{2} + 3 \tilde y^{2} + 3 \sqrt{6} \tilde z = 0$$
y
$$4 \tilde x^{2} + 3 \tilde y^{2} - 3 \sqrt{6} \tilde z = 0$$
$$2 \tilde z + \left(\frac{\tilde x^{2}}{\frac{3}{8} \sqrt{6}} + \frac{\tilde y^{2}}{\frac{1}{2} \sqrt{6}}\right) = 0$$
y
$$- 2 \tilde z + \left(\frac{\tilde x^{2}}{\frac{3}{8} \sqrt{6}} + \frac{\tilde y^{2}}{\frac{1}{2} \sqrt{6}}\right) = 0$$
es la ecuación para el tipo paraboloide elíptico
- está reducida a la forma canónica