4x^2+y^2+4z^2-4y-24z+36=0 forma canónica

Se da la ecuación de superficie de 2 grado:
$$4 x^{2} + y^{2} - 4 y + 4 z^{2} - 24 z + 36 = 0$$
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
$$a_{11} = 4$$
$$a_{12} = 0$$
$$a_{13} = 0$$
$$a_{14} = 0$$
$$a_{22} = 1$$
$$a_{23} = 0$$
$$a_{24} = -2$$
$$a_{33} = 4$$
$$a_{34} = -12$$
$$a_{44} = 36$$
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
$$I_{1} = 9$$

     |4  0|   |1  0|   |4  0|
I2 = |    | + |    | + |    |
     |0  1|   |0  4|   |0  4|

$$I_{3} = \left|\begin{matrix}4 & 0 & 0\\0 & 1 & 0\\0 & 0 & 4\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}4 & 0 & 0 & 0\\0 & 1 & 0 & -2\\0 & 0 & 4 & -12\\0 & -2 & -12 & 36\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}4 - \lambda & 0 & 0\\0 & 1 - \lambda & 0\\0 & 0 & 4 - \lambda\end{matrix}\right|$$

     |4  0 |   |1   -2|   | 4   -12|
K2 = |     | + |      | + |        |
     |0  36|   |-2  36|   |-12  36 |

     |4  0   0 |   |1    0   -2 |   |4   0    0 |
     |         |   |            |   |           |
K3 = |0  1   -2| + |0    4   -12| + |0   4   -12|
     |         |   |            |   |           |
     |0  -2  36|   |-2  -12  36 |   |0  -12  36 |

$$I_{1} = 9$$
$$I_{2} = 24$$
$$I_{3} = 16$$
$$I_{4} = -64$$
$$I{\left(\lambda \right)} = - \lambda^{3} + 9 \lambda^{2} - 24 \lambda + 16$$
$$K_{2} = 176$$
$$K_{3} = 112$$
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
$$\lambda^{3} - 9 \lambda^{2} + 24 \lambda - 16 = 0$$
$$\lambda_{1} = 1$$
$$\lambda_{2} = 4$$
$$\lambda_{3} = 4$$
entonces la forma canónica de la ecuación será
$$\left(\tilde z^{2} \lambda_{3} + \left(\tilde x^{2} \lambda_{1} + \tilde y^{2} \lambda_{2}\right)\right) + \frac{I_{4}}{I_{3}} = 0$$
$$\tilde x^{2} + 4 \tilde y^{2} + 4 \tilde z^{2} - 4 = 0$$

        2           2           2    
\tilde x    \tilde y    \tilde z     
--------- + --------- + --------- = 1
      2             2           2    
  / 1\       /  1  \     /  1  \     
  \2 /       |-----|     |-----|     
             \2*1/2/     \2*1/2/     

es la ecuación para el tipo elipsoide
- está reducida a la forma canónica