Sr Examen
36x^2+16y^2-9z^2-144x+96y+288=0 forma canónica
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Solución
Ha introducido
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2 2 2 288 - 144*x - 9*z + 16*y + 36*x + 96*y = 0
$$36 x^{2} - 144 x + 16 y^{2} + 96 y - 9 z^{2} + 288 = 0$$
36*x^2 - 144*x + 16*y^2 + 96*y - 9*z^2 + 288 = 0
Método de invariantes
Se da la ecuación de superficie de 2 grado:
$$36 x^{2} - 144 x + 16 y^{2} + 96 y - 9 z^{2} + 288 = 0$$
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
$$a_{11} = 36$$
$$a_{12} = 0$$
$$a_{13} = 0$$
$$a_{14} = -72$$
$$a_{22} = 16$$
$$a_{23} = 0$$
$$a_{24} = 48$$
$$a_{33} = -9$$
$$a_{34} = 0$$
$$a_{44} = 288$$
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
$$I_{1} = 43$$
$$I_{3} = \left|\begin{matrix}36 & 0 & 0\\0 & 16 & 0\\0 & 0 & -9\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}36 & 0 & 0 & -72\\0 & 16 & 0 & 48\\0 & 0 & -9 & 0\\-72 & 48 & 0 & 288\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}36 - \lambda & 0 & 0\\0 & 16 - \lambda & 0\\0 & 0 & - \lambda - 9\end{matrix}\right|$$
$$I_{1} = 43$$
$$I_{2} = 108$$
$$I_{3} = -5184$$
$$I_{4} = 0$$
$$I{\left(\lambda \right)} = - \lambda^{3} + 43 \lambda^{2} - 108 \lambda - 5184$$
$$K_{2} = 4896$$
$$K_{3} = -67392$$
Como
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
$$\lambda^{3} - 43 \lambda^{2} + 108 \lambda + 5184 = 0$$
$$\lambda_{1} = 36$$
$$\lambda_{2} = 16$$
$$\lambda_{3} = -9$$
entonces la forma canónica de la ecuación será
$$\left(\tilde z^{2} \lambda_{3} + \left(\tilde x^{2} \lambda_{1} + \tilde y^{2} \lambda_{2}\right)\right) + \frac{I_{4}}{I_{3}} = 0$$
$$36 \tilde x^{2} + 16 \tilde y^{2} - 9 \tilde z^{2} = 0$$
$$- \frac{\tilde z^{2}}{\left(\frac{1}{3}\right)^{2}} + \left(\frac{\tilde x^{2}}{\left(\frac{1}{6}\right)^{2}} + \frac{\tilde y^{2}}{\left(\frac{1}{4}\right)^{2}}\right) = 0$$
es la ecuación para el tipo cono
- está reducida a la forma canónica
$$36 x^{2} - 144 x + 16 y^{2} + 96 y - 9 z^{2} + 288 = 0$$
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
$$a_{11} = 36$$
$$a_{12} = 0$$
$$a_{13} = 0$$
$$a_{14} = -72$$
$$a_{22} = 16$$
$$a_{23} = 0$$
$$a_{24} = 48$$
$$a_{33} = -9$$
$$a_{34} = 0$$
$$a_{44} = 288$$
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
$$I_{1} = 43$$
|36 0 | |16 0 | |36 0 |
I2 = | | + | | + | |
|0 16| |0 -9| |0 -9|$$I_{3} = \left|\begin{matrix}36 & 0 & 0\\0 & 16 & 0\\0 & 0 & -9\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}36 & 0 & 0 & -72\\0 & 16 & 0 & 48\\0 & 0 & -9 & 0\\-72 & 48 & 0 & 288\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}36 - \lambda & 0 & 0\\0 & 16 - \lambda & 0\\0 & 0 & - \lambda - 9\end{matrix}\right|$$
|36 -72| |16 48 | |-9 0 |
K2 = | | + | | + | |
|-72 288| |48 288| |0 288| |36 0 -72| |16 0 48 | |36 0 -72|
| | | | | |
K3 = | 0 16 48 | + |0 -9 0 | + | 0 -9 0 |
| | | | | |
|-72 48 288| |48 0 288| |-72 0 288|$$I_{1} = 43$$
$$I_{2} = 108$$
$$I_{3} = -5184$$
$$I_{4} = 0$$
$$I{\left(\lambda \right)} = - \lambda^{3} + 43 \lambda^{2} - 108 \lambda - 5184$$
$$K_{2} = 4896$$
$$K_{3} = -67392$$
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
$$\lambda^{3} - 43 \lambda^{2} + 108 \lambda + 5184 = 0$$
$$\lambda_{1} = 36$$
$$\lambda_{2} = 16$$
$$\lambda_{3} = -9$$
entonces la forma canónica de la ecuación será
$$\left(\tilde z^{2} \lambda_{3} + \left(\tilde x^{2} \lambda_{1} + \tilde y^{2} \lambda_{2}\right)\right) + \frac{I_{4}}{I_{3}} = 0$$
$$36 \tilde x^{2} + 16 \tilde y^{2} - 9 \tilde z^{2} = 0$$
$$- \frac{\tilde z^{2}}{\left(\frac{1}{3}\right)^{2}} + \left(\frac{\tilde x^{2}}{\left(\frac{1}{6}\right)^{2}} + \frac{\tilde y^{2}}{\left(\frac{1}{4}\right)^{2}}\right) = 0$$
es la ecuación para el tipo cono
- está reducida a la forma canónica