((y^2)/3)+((y^2)/4)-((z^2)/25)=-1 forma canónica

Se da la ecuación de superficie de 2 grado:
$$\frac{7 y^{2}}{12} - \frac{z^{2}}{25} + 1 = 0$$
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
$$a_{11} = 0$$
$$a_{12} = 0$$
$$a_{13} = 0$$
$$a_{14} = 0$$
$$a_{22} = \frac{7}{12}$$
$$a_{23} = 0$$
$$a_{24} = 0$$
$$a_{33} = - \frac{1}{25}$$
$$a_{34} = 0$$
$$a_{44} = 1$$
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
$$I_{1} = \frac{163}{300}$$

     |0   0  |   |7/12    0  |   |0    0  |
I2 = |       | + |           | + |        |
     |0  7/12|   | 0    -1/25|   |0  -1/25|

$$I_{3} = \left|\begin{matrix}0 & 0 & 0\\0 & \frac{7}{12} & 0\\0 & 0 & - \frac{1}{25}\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}0 & 0 & 0 & 0\\0 & \frac{7}{12} & 0 & 0\\0 & 0 & - \frac{1}{25} & 0\\0 & 0 & 0 & 1\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda & 0 & 0\\0 & \frac{7}{12} - \lambda & 0\\0 & 0 & - \lambda - \frac{1}{25}\end{matrix}\right|$$

     |0  0|   |7/12  0|   |-1/25  0|
K2 = |    | + |       | + |        |
     |0  1|   | 0    1|   |  0    1|

     |0   0    0|   |7/12    0    0|   |0    0    0|
     |          |   |              |   |           |
K3 = |0  7/12  0| + | 0    -1/25  0| + |0  -1/25  0|
     |          |   |              |   |           |
     |0   0    1|   | 0      0    1|   |0    0    1|

$$I_{1} = \frac{163}{300}$$
$$I_{2} = - \frac{7}{300}$$
$$I_{3} = 0$$
$$I_{4} = 0$$
$$I{\left(\lambda \right)} = - \lambda^{3} + \frac{163 \lambda^{2}}{300} + \frac{7 \lambda}{300}$$
$$K_{2} = \frac{163}{300}$$
$$K_{3} = - \frac{7}{300}$$
Como
$$I_{3} = 0 \wedge I_{4} = 0 \wedge I_{2} \neq 0$$
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
$$\lambda^{3} - \frac{163 \lambda^{2}}{300} - \frac{7 \lambda}{300} = 0$$
$$\lambda_{1} = \frac{7}{12}$$
$$\lambda_{2} = - \frac{1}{25}$$
$$\lambda_{3} = 0$$
entonces la forma canónica de la ecuación será
$$\left(\tilde x^{2} \lambda_{1} + \tilde y^{2} \lambda_{2}\right) + \frac{K_{3}}{I_{2}} = 0$$
$$\frac{7 \tilde x^{2}}{12} - \frac{\tilde y^{2}}{25} + 1 = 0$$
$$\frac{\tilde x^{2}}{\frac{12}{7}} - \frac{\tilde y^{2}}{25} = -1$$
es la ecuación para el tipo cilindro hiperbólico
- está reducida a la forma canónica