(x^2/1)-(y^2/9)-(z^2/16)=1 forma canónica

Se da la ecuación de superficie de 2 grado:
$$x^{2} - \frac{y^{2}}{9} - \frac{z^{2}}{16} - 1 = 0$$
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
$$a_{11} = 1$$
$$a_{12} = 0$$
$$a_{13} = 0$$
$$a_{14} = 0$$
$$a_{22} = - \frac{1}{9}$$
$$a_{23} = 0$$
$$a_{24} = 0$$
$$a_{33} = - \frac{1}{16}$$
$$a_{34} = 0$$
$$a_{44} = -1$$
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
$$I_{1} = \frac{119}{144}$$

     |1   0  |   |-1/9    0  |   |1    0  |
I2 = |       | + |           | + |        |
     |0  -1/9|   | 0    -1/16|   |0  -1/16|

$$I_{3} = \left|\begin{matrix}1 & 0 & 0\\0 & - \frac{1}{9} & 0\\0 & 0 & - \frac{1}{16}\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}1 & 0 & 0 & 0\\0 & - \frac{1}{9} & 0 & 0\\0 & 0 & - \frac{1}{16} & 0\\0 & 0 & 0 & -1\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}1 - \lambda & 0 & 0\\0 & - \lambda - \frac{1}{9} & 0\\0 & 0 & - \lambda - \frac{1}{16}\end{matrix}\right|$$

     |1  0 |   |-1/9  0 |   |-1/16  0 |
K2 = |     | + |        | + |         |
     |0  -1|   | 0    -1|   |  0    -1|

     |1   0    0 |   |-1/9    0    0 |   |1    0    0 |
     |           |   |               |   |            |
K3 = |0  -1/9  0 | + | 0    -1/16  0 | + |0  -1/16  0 |
     |           |   |               |   |            |
     |0   0    -1|   | 0      0    -1|   |0    0    -1|

$$I_{1} = \frac{119}{144}$$
$$I_{2} = - \frac{1}{6}$$
$$I_{3} = \frac{1}{144}$$
$$I_{4} = - \frac{1}{144}$$
$$I{\left(\lambda \right)} = - \lambda^{3} + \frac{119 \lambda^{2}}{144} + \frac{\lambda}{6} + \frac{1}{144}$$
$$K_{2} = - \frac{119}{144}$$
$$K_{3} = \frac{1}{6}$$
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
$$\lambda^{3} - \frac{119 \lambda^{2}}{144} - \frac{\lambda}{6} - \frac{1}{144} = 0$$
$$\lambda_{1} = 1$$
$$\lambda_{2} = - \frac{1}{9}$$
$$\lambda_{3} = - \frac{1}{16}$$
entonces la forma canónica de la ecuación será
$$\left(\tilde z^{2} \lambda_{3} + \left(\tilde x^{2} \lambda_{1} + \tilde y^{2} \lambda_{2}\right)\right) + \frac{I_{4}}{I_{3}} = 0$$
$$\tilde x^{2} - \frac{\tilde y^{2}}{9} - \frac{\tilde z^{2}}{16} - 1 = 0$$
$$- \frac{\tilde x^{2}}{\left(1^{-1}\right)^{2}} + \left(\frac{\tilde y^{2}}{\left(\frac{3}{1}\right)^{2}} + \frac{\tilde z^{2}}{\left(\frac{4}{1}\right)^{2}}\right) = -1$$
es la ecuación para el tipo hiperboloide bilateral
- está reducida a la forma canónica