Sr Examen
-2x^2+8xy+y^2-2xy-8z+x^2+y^2=0 forma canónica
El profesor se sorprenderá mucho al ver tu solución correcta😉
Solución
Ha introducido
[src]
2 2 - x - 8*z + 2*y + 6*x*y = 0
$$- x^{2} + 6 x y + 2 y^{2} - 8 z = 0$$
-x^2 + 6*x*y + 2*y^2 - 8*z = 0
Método de invariantes
Se da la ecuación de superficie de 2 grado:
$$- x^{2} + 6 x y + 2 y^{2} - 8 z = 0$$
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
$$a_{11} = -1$$
$$a_{12} = 3$$
$$a_{13} = 0$$
$$a_{14} = 0$$
$$a_{22} = 2$$
$$a_{23} = 0$$
$$a_{24} = 0$$
$$a_{33} = 0$$
$$a_{34} = -4$$
$$a_{44} = 0$$
Las invariantes de esta ecuación al transformar las coordenadas son los determinantes:
$$I_{1} = a_{11} + a_{22} + a_{33}$$
$$I_{3} = \left|\begin{matrix}a_{11} & a_{12} & a_{13}\\a_{12} & a_{22} & a_{23}\\a_{13} & a_{23} & a_{33}\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}a_{11} & a_{12} & a_{13} & a_{14}\\a_{12} & a_{22} & a_{23} & a_{24}\\a_{13} & a_{23} & a_{33} & a_{34}\\a_{14} & a_{24} & a_{34} & a_{44}\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}a_{11} - \lambda & a_{12} & a_{13}\\a_{12} & a_{22} - \lambda & a_{23}\\a_{13} & a_{23} & a_{33} - \lambda\end{matrix}\right|$$
sustituimos coeficientes
$$I_{1} = 1$$
$$I_{3} = \left|\begin{matrix}-1 & 3 & 0\\3 & 2 & 0\\0 & 0 & 0\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}-1 & 3 & 0 & 0\\3 & 2 & 0 & 0\\0 & 0 & 0 & -4\\0 & 0 & -4 & 0\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda - 1 & 3 & 0\\3 & 2 - \lambda & 0\\0 & 0 & - \lambda\end{matrix}\right|$$
$$I_{1} = 1$$
$$I_{2} = -11$$
$$I_{3} = 0$$
$$I_{4} = 176$$
$$I{\left(\lambda \right)} = - \lambda^{3} + \lambda^{2} + 11 \lambda$$
$$K_{2} = -16$$
$$K_{3} = -16$$
Como
$$I_{3} = 0 \wedge I_{2} \neq 0 \wedge I_{4} \neq 0$$
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
$$\lambda^{3} - \lambda^{2} - 11 \lambda = 0$$
$$\lambda_{1} = \frac{1}{2} - \frac{3 \sqrt{5}}{2}$$
$$\lambda_{2} = \frac{1}{2} + \frac{3 \sqrt{5}}{2}$$
$$\lambda_{3} = 0$$
entonces la forma canónica de la ecuación será
$$\tilde z 2 \sqrt{\frac{\left(-1\right) I_{4}}{I_{2}}} + \left(\tilde x^{2} \lambda_{1} + \tilde y^{2} \lambda_{2}\right) = 0$$
y
$$- \tilde z 2 \sqrt{\frac{\left(-1\right) I_{4}}{I_{2}}} + \left(\tilde x^{2} \lambda_{1} + \tilde y^{2} \lambda_{2}\right) = 0$$
$$\tilde x^{2} \left(\frac{1}{2} - \frac{3 \sqrt{5}}{2}\right) + \tilde y^{2} \left(\frac{1}{2} + \frac{3 \sqrt{5}}{2}\right) + 8 \tilde z = 0$$
y
$$\tilde x^{2} \left(\frac{1}{2} - \frac{3 \sqrt{5}}{2}\right) + \tilde y^{2} \left(\frac{1}{2} + \frac{3 \sqrt{5}}{2}\right) - 8 \tilde z = 0$$
$$- 2 \tilde z + \left(\frac{\tilde x^{2}}{4 \frac{1}{- \frac{1}{2} + \frac{3 \sqrt{5}}{2}}} - \frac{\tilde y^{2}}{4 \frac{1}{\frac{1}{2} + \frac{3 \sqrt{5}}{2}}}\right) = 0$$
y
$$2 \tilde z + \left(\frac{\tilde x^{2}}{4 \frac{1}{- \frac{1}{2} + \frac{3 \sqrt{5}}{2}}} - \frac{\tilde y^{2}}{4 \frac{1}{\frac{1}{2} + \frac{3 \sqrt{5}}{2}}}\right) = 0$$
es la ecuación para el tipo paraboloide hiperbólico
- está reducida a la forma canónica
$$- x^{2} + 6 x y + 2 y^{2} - 8 z = 0$$
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
$$a_{11} = -1$$
$$a_{12} = 3$$
$$a_{13} = 0$$
$$a_{14} = 0$$
$$a_{22} = 2$$
$$a_{23} = 0$$
$$a_{24} = 0$$
$$a_{33} = 0$$
$$a_{34} = -4$$
$$a_{44} = 0$$
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
$$I_{1} = 1$$
|-1 3| |2 0| |-1 0|
I2 = | | + | | + | |
|3 2| |0 0| |0 0|$$I_{3} = \left|\begin{matrix}-1 & 3 & 0\\3 & 2 & 0\\0 & 0 & 0\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}-1 & 3 & 0 & 0\\3 & 2 & 0 & 0\\0 & 0 & 0 & -4\\0 & 0 & -4 & 0\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda - 1 & 3 & 0\\3 & 2 - \lambda & 0\\0 & 0 & - \lambda\end{matrix}\right|$$
|-1 0| |2 0| |0 -4|
K2 = | | + | | + | |
|0 0| |0 0| |-4 0 | |-1 3 0| |2 0 0 | |-1 0 0 |
| | | | | |
K3 = |3 2 0| + |0 0 -4| + |0 0 -4|
| | | | | |
|0 0 0| |0 -4 0 | |0 -4 0 |$$I_{1} = 1$$
$$I_{2} = -11$$
$$I_{3} = 0$$
$$I_{4} = 176$$
$$I{\left(\lambda \right)} = - \lambda^{3} + \lambda^{2} + 11 \lambda$$
$$K_{2} = -16$$
$$K_{3} = -16$$
Como
$$I_{3} = 0 \wedge I_{2} \neq 0 \wedge I_{4} \neq 0$$
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
$$\lambda^{3} - \lambda^{2} - 11 \lambda = 0$$
$$\lambda_{1} = \frac{1}{2} - \frac{3 \sqrt{5}}{2}$$
$$\lambda_{2} = \frac{1}{2} + \frac{3 \sqrt{5}}{2}$$
$$\lambda_{3} = 0$$
entonces la forma canónica de la ecuación será
$$\tilde z 2 \sqrt{\frac{\left(-1\right) I_{4}}{I_{2}}} + \left(\tilde x^{2} \lambda_{1} + \tilde y^{2} \lambda_{2}\right) = 0$$
y
$$- \tilde z 2 \sqrt{\frac{\left(-1\right) I_{4}}{I_{2}}} + \left(\tilde x^{2} \lambda_{1} + \tilde y^{2} \lambda_{2}\right) = 0$$
$$\tilde x^{2} \left(\frac{1}{2} - \frac{3 \sqrt{5}}{2}\right) + \tilde y^{2} \left(\frac{1}{2} + \frac{3 \sqrt{5}}{2}\right) + 8 \tilde z = 0$$
y
$$\tilde x^{2} \left(\frac{1}{2} - \frac{3 \sqrt{5}}{2}\right) + \tilde y^{2} \left(\frac{1}{2} + \frac{3 \sqrt{5}}{2}\right) - 8 \tilde z = 0$$
$$- 2 \tilde z + \left(\frac{\tilde x^{2}}{4 \frac{1}{- \frac{1}{2} + \frac{3 \sqrt{5}}{2}}} - \frac{\tilde y^{2}}{4 \frac{1}{\frac{1}{2} + \frac{3 \sqrt{5}}{2}}}\right) = 0$$
y
$$2 \tilde z + \left(\frac{\tilde x^{2}}{4 \frac{1}{- \frac{1}{2} + \frac{3 \sqrt{5}}{2}}} - \frac{\tilde y^{2}}{4 \frac{1}{\frac{1}{2} + \frac{3 \sqrt{5}}{2}}}\right) = 0$$
es la ecuación para el tipo paraboloide hiperbólico
- está reducida a la forma canónica