x^2-4y^2+9z^2-36=0 forma canónica

Se da la ecuación de superficie de 2 grado:
$$x^{2} - 4 y^{2} + 9 z^{2} - 36 = 0$$
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
$$a_{11} = 1$$
$$a_{12} = 0$$
$$a_{13} = 0$$
$$a_{14} = 0$$
$$a_{22} = -4$$
$$a_{23} = 0$$
$$a_{24} = 0$$
$$a_{33} = 9$$
$$a_{34} = 0$$
$$a_{44} = -36$$
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
$$I_{1} = 6$$

     |1  0 |   |-4  0|   |1  0|
I2 = |     | + |     | + |    |
     |0  -4|   |0   9|   |0  9|

$$I_{3} = \left|\begin{matrix}1 & 0 & 0\\0 & -4 & 0\\0 & 0 & 9\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}1 & 0 & 0 & 0\\0 & -4 & 0 & 0\\0 & 0 & 9 & 0\\0 & 0 & 0 & -36\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}1 - \lambda & 0 & 0\\0 & - \lambda - 4 & 0\\0 & 0 & 9 - \lambda\end{matrix}\right|$$

     |1   0 |   |-4   0 |   |9   0 |
K2 = |      | + |       | + |      |
     |0  -36|   |0   -36|   |0  -36|

     |1  0    0 |   |-4  0   0 |   |1  0   0 |
     |          |   |          |   |         |
K3 = |0  -4   0 | + |0   9   0 | + |0  9   0 |
     |          |   |          |   |         |
     |0  0   -36|   |0   0  -36|   |0  0  -36|

$$I_{1} = 6$$
$$I_{2} = -31$$
$$I_{3} = -36$$
$$I_{4} = 1296$$
$$I{\left(\lambda \right)} = - \lambda^{3} + 6 \lambda^{2} + 31 \lambda - 36$$
$$K_{2} = -216$$
$$K_{3} = 1116$$
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
$$\lambda^{3} - 6 \lambda^{2} - 31 \lambda + 36 = 0$$
$$\lambda_{1} = 9$$
$$\lambda_{2} = 1$$
$$\lambda_{3} = -4$$
entonces la forma canónica de la ecuación será
$$\left(\tilde z^{2} \lambda_{3} + \left(\tilde x^{2} \lambda_{1} + \tilde y^{2} \lambda_{2}\right)\right) + \frac{I_{4}}{I_{3}} = 0$$
$$9 \tilde x^{2} + \tilde y^{2} - 4 \tilde z^{2} - 36 = 0$$

        2           2           2    
\tilde x    \tilde y    \tilde z     
--------- + --------- - --------- = 1
        2         2             2    
 /  1  \      / 1\       /  1  \     
 |-----|      \6 /       |-----|     
 \3*1/6/                 \2*1/6/     

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