Sr Examen
11x^2+9y^2+0z^2-6xy-2x+6y+20z+11=0 forma canónica
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Solución
Ha introducido
[src]
2 2 11 - 2*x + 6*y + 9*y + 11*x + 20*z - 6*x*y = 0
$$11 x^{2} - 6 x y - 2 x + 9 y^{2} + 6 y + 20 z + 11 = 0$$
11*x^2 - 6*x*y - 2*x + 9*y^2 + 6*y + 20*z + 11 = 0
Método de invariantes
Se da la ecuación de superficie de 2 grado:
$$11 x^{2} - 6 x y - 2 x + 9 y^{2} + 6 y + 20 z + 11 = 0$$
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
$$a_{11} = 11$$
$$a_{12} = -3$$
$$a_{13} = 0$$
$$a_{14} = -1$$
$$a_{22} = 9$$
$$a_{23} = 0$$
$$a_{24} = 3$$
$$a_{33} = 0$$
$$a_{34} = 10$$
$$a_{44} = 11$$
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
$$I_{1} = 20$$
$$I_{3} = \left|\begin{matrix}11 & -3 & 0\\-3 & 9 & 0\\0 & 0 & 0\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}11 & -3 & 0 & -1\\-3 & 9 & 0 & 3\\0 & 0 & 0 & 10\\-1 & 3 & 10 & 11\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}11 - \lambda & -3 & 0\\-3 & 9 - \lambda & 0\\0 & 0 & - \lambda\end{matrix}\right|$$
$$I_{1} = 20$$
$$I_{2} = 90$$
$$I_{3} = 0$$
$$I_{4} = -9000$$
$$I{\left(\lambda \right)} = - \lambda^{3} + 20 \lambda^{2} - 90 \lambda$$
$$K_{2} = 110$$
$$K_{3} = -1100$$
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
$$\lambda^{3} - 20 \lambda^{2} + 90 \lambda = 0$$
$$\lambda_{1} = 10 - \sqrt{10}$$
$$\lambda_{2} = \sqrt{10} + 10$$
$$\lambda_{3} = 0$$
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
$$- \tilde z 2 \sqrt{\frac{\left(-1\right) I_{4}}{I_{2}}} + \left(\tilde x^{2} \lambda_{1} + \tilde y^{2} \lambda_{2}\right) = 0$$
$$\tilde x^{2} \left(10 - \sqrt{10}\right) + \tilde y^{2} \left(\sqrt{10} + 10\right) + 20 \tilde z = 0$$
y
$$\tilde x^{2} \left(10 - \sqrt{10}\right) + \tilde y^{2} \left(\sqrt{10} + 10\right) - 20 \tilde z = 0$$
$$2 \tilde z + \left(\frac{\tilde x^{2}}{10 \frac{1}{10 - \sqrt{10}}} + \frac{\tilde y^{2}}{10 \frac{1}{\sqrt{10} + 10}}\right) = 0$$
y
$$- 2 \tilde z + \left(\frac{\tilde x^{2}}{10 \frac{1}{10 - \sqrt{10}}} + \frac{\tilde y^{2}}{10 \frac{1}{\sqrt{10} + 10}}\right) = 0$$
es la ecuación para el tipo paraboloide elíptico
- está reducida a la forma canónica
$$11 x^{2} - 6 x y - 2 x + 9 y^{2} + 6 y + 20 z + 11 = 0$$
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
$$a_{11} = 11$$
$$a_{12} = -3$$
$$a_{13} = 0$$
$$a_{14} = -1$$
$$a_{22} = 9$$
$$a_{23} = 0$$
$$a_{24} = 3$$
$$a_{33} = 0$$
$$a_{34} = 10$$
$$a_{44} = 11$$
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
$$I_{1} = 20$$
|11 -3| |9 0| |11 0|
I2 = | | + | | + | |
|-3 9 | |0 0| |0 0|$$I_{3} = \left|\begin{matrix}11 & -3 & 0\\-3 & 9 & 0\\0 & 0 & 0\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}11 & -3 & 0 & -1\\-3 & 9 & 0 & 3\\0 & 0 & 0 & 10\\-1 & 3 & 10 & 11\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}11 - \lambda & -3 & 0\\-3 & 9 - \lambda & 0\\0 & 0 & - \lambda\end{matrix}\right|$$
|11 -1| |9 3 | |0 10|
K2 = | | + | | + | |
|-1 11| |3 11| |10 11| |11 -3 -1| |9 0 3 | |11 0 -1|
| | | | | |
K3 = |-3 9 3 | + |0 0 10| + |0 0 10|
| | | | | |
|-1 3 11| |3 10 11| |-1 10 11|$$I_{1} = 20$$
$$I_{2} = 90$$
$$I_{3} = 0$$
$$I_{4} = -9000$$
$$I{\left(\lambda \right)} = - \lambda^{3} + 20 \lambda^{2} - 90 \lambda$$
$$K_{2} = 110$$
$$K_{3} = -1100$$
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
$$\lambda^{3} - 20 \lambda^{2} + 90 \lambda = 0$$
$$\lambda_{1} = 10 - \sqrt{10}$$
$$\lambda_{2} = \sqrt{10} + 10$$
$$\lambda_{3} = 0$$
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
$$- \tilde z 2 \sqrt{\frac{\left(-1\right) I_{4}}{I_{2}}} + \left(\tilde x^{2} \lambda_{1} + \tilde y^{2} \lambda_{2}\right) = 0$$
$$\tilde x^{2} \left(10 - \sqrt{10}\right) + \tilde y^{2} \left(\sqrt{10} + 10\right) + 20 \tilde z = 0$$
y
$$\tilde x^{2} \left(10 - \sqrt{10}\right) + \tilde y^{2} \left(\sqrt{10} + 10\right) - 20 \tilde z = 0$$
$$2 \tilde z + \left(\frac{\tilde x^{2}}{10 \frac{1}{10 - \sqrt{10}}} + \frac{\tilde y^{2}}{10 \frac{1}{\sqrt{10} + 10}}\right) = 0$$
y
$$- 2 \tilde z + \left(\frac{\tilde x^{2}}{10 \frac{1}{10 - \sqrt{10}}} + \frac{\tilde y^{2}}{10 \frac{1}{\sqrt{10} + 10}}\right) = 0$$
es la ecuación para el tipo paraboloide elíptico
- está reducida a la forma canónica