Se da la ecuación de superficie de 2 grado:
$$5 y^{2} + 14 y z - 2 y - 3 z^{2} + 10 z - 3 = 0$$
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
$$a_{11} = 0$$
$$a_{12} = 0$$
$$a_{13} = 0$$
$$a_{14} = 0$$
$$a_{22} = 5$$
$$a_{23} = 7$$
$$a_{24} = -1$$
$$a_{33} = -3$$
$$a_{34} = 5$$
$$a_{44} = -3$$
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
$$I_{1} = 2$$
|0 0| |5 7 | |0 0 |
I2 = | | + | | + | |
|0 5| |7 -3| |0 -3|
$$I_{3} = \left|\begin{matrix}0 & 0 & 0\\0 & 5 & 7\\0 & 7 & -3\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}0 & 0 & 0 & 0\\0 & 5 & 7 & -1\\0 & 7 & -3 & 5\\0 & -1 & 5 & -3\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda & 0 & 0\\0 & 5 - \lambda & 7\\0 & 7 & - \lambda - 3\end{matrix}\right|$$
|0 0 | |5 -1| |-3 5 |
K2 = | | + | | + | |
|0 -3| |-1 -3| |5 -3|
|0 0 0 | |5 7 -1| |0 0 0 |
| | | | | |
K3 = |0 5 -1| + |7 -3 5 | + |0 -3 5 |
| | | | | |
|0 -1 -3| |-1 5 -3| |0 5 -3|
$$I_{1} = 2$$
$$I_{2} = -64$$
$$I_{3} = 0$$
$$I_{4} = 0$$
$$I{\left(\lambda \right)} = - \lambda^{3} + 2 \lambda^{2} + 64 \lambda$$
$$K_{2} = -32$$
$$K_{3} = 0$$
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
$$\lambda^{3} - 2 \lambda^{2} - 64 \lambda = 0$$
$$\lambda_{1} = 1 - \sqrt{65}$$
$$\lambda_{2} = 1 + \sqrt{65}$$
$$\lambda_{3} = 0$$
entonces la forma canónica de la ecuación será
$$\left(\tilde x^{2} \lambda_{1} + \tilde y^{2} \lambda_{2}\right) + \frac{K_{3}}{I_{2}} = 0$$
$$\tilde x^{2} \left(1 - \sqrt{65}\right) + \tilde y^{2} \left(1 + \sqrt{65}\right) = 0$$
$$\frac{\tilde x^{2}}{\left(\frac{1}{\sqrt{-1 + \sqrt{65}}}\right)^{2}} - \frac{\tilde y^{2}}{\left(\frac{1}{\sqrt{1 + \sqrt{65}}}\right)^{2}} = 0$$
es la ecuación para el tipo dos planos intersectantes
- está reducida a la forma canónica