8^2+5z^2+5*x^2-4*x*y+8*x*z+4*y*z+6*y+6*z-6*x+10=0 forma canónica

Se da la ecuación de superficie de 2 grado:
$$5 x^{2} - 4 x y + 8 x z - 6 x + 4 y z + 6 y + 5 z^{2} + 6 z + 74 = 0$$
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
$$a_{11} = 5$$
$$a_{12} = -2$$
$$a_{13} = 4$$
$$a_{14} = -3$$
$$a_{22} = 0$$
$$a_{23} = 2$$
$$a_{24} = 3$$
$$a_{33} = 5$$
$$a_{34} = 3$$
$$a_{44} = 74$$
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
$$I_{1} = 10$$

     |5   -2|   |0  2|   |5  4|
I2 = |      | + |    | + |    |
     |-2  0 |   |2  5|   |4  5|

$$I_{3} = \left|\begin{matrix}5 & -2 & 4\\-2 & 0 & 2\\4 & 2 & 5\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}5 & -2 & 4 & -3\\-2 & 0 & 2 & 3\\4 & 2 & 5 & 3\\-3 & 3 & 3 & 74\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}5 - \lambda & -2 & 4\\-2 & - \lambda & 2\\4 & 2 & 5 - \lambda\end{matrix}\right|$$

     |5   -3|   |0  3 |   |5  3 |
K2 = |      | + |     | + |     |
     |-3  74|   |3  74|   |3  74|

     |5   -2  -3|   |0  2  3 |   |5   4  -3|
     |          |   |        |   |         |
K3 = |-2  0   3 | + |2  5  3 | + |4   5  3 |
     |          |   |        |   |         |
     |-3  3   74|   |3  3  74|   |-3  3  74|

$$I_{1} = 10$$
$$I_{2} = 1$$
$$I_{3} = -72$$
$$I_{4} = -4761$$
$$I{\left(\lambda \right)} = - \lambda^{3} + 10 \lambda^{2} - \lambda - 72$$
$$K_{2} = 713$$
$$K_{3} = -106$$
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
$$\lambda^{3} - 10 \lambda^{2} + \lambda + 72 = 0$$
$$\lambda_{1} = 9$$
$$\lambda_{2} = \frac{1}{2} - \frac{\sqrt{33}}{2}$$
$$\lambda_{3} = \frac{1}{2} + \frac{\sqrt{33}}{2}$$
entonces la forma canónica de la ecuación será
$$\left(\tilde z^{2} \lambda_{3} + \left(\tilde x^{2} \lambda_{1} + \tilde y^{2} \lambda_{2}\right)\right) + \frac{I_{4}}{I_{3}} = 0$$
$$9 \tilde x^{2} + \tilde y^{2} \left(\frac{1}{2} - \frac{\sqrt{33}}{2}\right) + \tilde z^{2} \left(\frac{1}{2} + \frac{\sqrt{33}}{2}\right) + \frac{529}{8} = 0$$
$$- \frac{\tilde y^{2}}{\left(\frac{1}{\frac{2 \sqrt{2}}{23} \sqrt{- \frac{1}{2} + \frac{\sqrt{33}}{2}}}\right)^{2}} + \left(\frac{\tilde x^{2}}{\left(\frac{1}{3 \frac{2 \sqrt{2}}{23}}\right)^{2}} + \frac{\tilde z^{2}}{\left(\frac{1}{\frac{2 \sqrt{2}}{23} \sqrt{\frac{1}{2} + \frac{\sqrt{33}}{2}}}\right)^{2}}\right) = -1$$
es la ecuación para el tipo hiperboloide bilateral
- está reducida a la forma canónica