3x^2-5y^2+2z^2-6=0 forma canónica

Se da la ecuación de superficie de 2 grado:
$$3 x^{2} - 5 y^{2} + 2 z^{2} - 6 = 0$$
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
$$a_{11} = 3$$
$$a_{12} = 0$$
$$a_{13} = 0$$
$$a_{14} = 0$$
$$a_{22} = -5$$
$$a_{23} = 0$$
$$a_{24} = 0$$
$$a_{33} = 2$$
$$a_{34} = 0$$
$$a_{44} = -6$$
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
$$I_{1} = 0$$

     |3  0 |   |-5  0|   |3  0|
I2 = |     | + |     | + |    |
     |0  -5|   |0   2|   |0  2|

$$I_{3} = \left|\begin{matrix}3 & 0 & 0\\0 & -5 & 0\\0 & 0 & 2\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}3 & 0 & 0 & 0\\0 & -5 & 0 & 0\\0 & 0 & 2 & 0\\0 & 0 & 0 & -6\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}3 - \lambda & 0 & 0\\0 & - \lambda - 5 & 0\\0 & 0 & 2 - \lambda\end{matrix}\right|$$

     |3  0 |   |-5  0 |   |2  0 |
K2 = |     | + |      | + |     |
     |0  -6|   |0   -6|   |0  -6|

     |3  0   0 |   |-5  0  0 |   |3  0  0 |
     |         |   |         |   |        |
K3 = |0  -5  0 | + |0   2  0 | + |0  2  0 |
     |         |   |         |   |        |
     |0  0   -6|   |0   0  -6|   |0  0  -6|

$$I_{1} = 0$$
$$I_{2} = -19$$
$$I_{3} = -30$$
$$I_{4} = 180$$
$$I{\left(\lambda \right)} = - \lambda^{3} + 19 \lambda - 30$$
$$K_{2} = 0$$
$$K_{3} = 114$$
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
$$\lambda^{3} - 19 \lambda + 30 = 0$$
$$\lambda_{1} = 3$$
$$\lambda_{2} = 2$$
$$\lambda_{3} = -5$$
entonces la forma canónica de la ecuación será
$$\left(\tilde z^{2} \lambda_{3} + \left(\tilde x^{2} \lambda_{1} + \tilde y^{2} \lambda_{2}\right)\right) + \frac{I_{4}}{I_{3}} = 0$$
$$3 \tilde x^{2} + 2 \tilde y^{2} - 5 \tilde z^{2} - 6 = 0$$
$$- \frac{\tilde z^{2}}{\left(\frac{\frac{1}{5} \sqrt{5}}{\frac{1}{6} \sqrt{6}}\right)^{2}} + \left(\frac{\tilde x^{2}}{\left(\frac{\frac{1}{3} \sqrt{3}}{\frac{1}{6} \sqrt{6}}\right)^{2}} + \frac{\tilde y^{2}}{\left(\frac{\frac{1}{2} \sqrt{2}}{\frac{1}{6} \sqrt{6}}\right)^{2}}\right) = 1$$
es la ecuación para el tipo hiperboloide unilateral
- está reducida a la forma canónica