Sr Examen
((x²)/25)+((y²)/16)-((z²)/4)-1=0 forma canónica
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Solución
Ha introducido
[src]
2 2 2
z y x
-1 - -- + -- + -- = 0
4 16 25
$$\frac{x^{2}}{25} + \frac{y^{2}}{16} - \frac{z^{2}}{4} - 1 = 0$$
x^2/25 + y^2/16 - z^2/4 - 1 = 0
Método de invariantes
Se da la ecuación de superficie de 2 grado:
$$\frac{x^{2}}{25} + \frac{y^{2}}{16} - \frac{z^{2}}{4} - 1 = 0$$
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
$$a_{11} = \frac{1}{25}$$
$$a_{12} = 0$$
$$a_{13} = 0$$
$$a_{14} = 0$$
$$a_{22} = \frac{1}{16}$$
$$a_{23} = 0$$
$$a_{24} = 0$$
$$a_{33} = - \frac{1}{4}$$
$$a_{34} = 0$$
$$a_{44} = -1$$
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
$$I_{1} = - \frac{59}{400}$$
$$I_{3} = \left|\begin{matrix}\frac{1}{25} & 0 & 0\\0 & \frac{1}{16} & 0\\0 & 0 & - \frac{1}{4}\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}\frac{1}{25} & 0 & 0 & 0\\0 & \frac{1}{16} & 0 & 0\\0 & 0 & - \frac{1}{4} & 0\\0 & 0 & 0 & -1\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}\frac{1}{25} - \lambda & 0 & 0\\0 & \frac{1}{16} - \lambda & 0\\0 & 0 & - \lambda - \frac{1}{4}\end{matrix}\right|$$
$$I_{1} = - \frac{59}{400}$$
$$I_{2} = - \frac{37}{1600}$$
$$I_{3} = - \frac{1}{1600}$$
$$I_{4} = \frac{1}{1600}$$
$$I{\left(\lambda \right)} = - \lambda^{3} - \frac{59 \lambda^{2}}{400} + \frac{37 \lambda}{1600} - \frac{1}{1600}$$
$$K_{2} = \frac{59}{400}$$
$$K_{3} = \frac{37}{1600}$$
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
$$\lambda^{3} + \frac{59 \lambda^{2}}{400} - \frac{37 \lambda}{1600} + \frac{1}{1600} = 0$$
$$\lambda_{1} = - \frac{1}{4}$$
$$\lambda_{2} = \frac{1}{16}$$
$$\lambda_{3} = \frac{1}{25}$$
entonces la forma canónica de la ecuación será
$$\left(\tilde z^{2} \lambda_{3} + \left(\tilde x^{2} \lambda_{1} + \tilde y^{2} \lambda_{2}\right)\right) + \frac{I_{4}}{I_{3}} = 0$$
$$- \frac{\tilde x^{2}}{4} + \frac{\tilde y^{2}}{16} + \frac{\tilde z^{2}}{25} - 1 = 0$$
$$- \frac{\tilde x^{2}}{\left(\frac{2}{1}\right)^{2}} + \left(\frac{\tilde y^{2}}{\left(\frac{4}{1}\right)^{2}} + \frac{\tilde z^{2}}{\left(\frac{5}{1}\right)^{2}}\right) = 1$$
es la ecuación para el tipo hiperboloide unilateral
- está reducida a la forma canónica
$$\frac{x^{2}}{25} + \frac{y^{2}}{16} - \frac{z^{2}}{4} - 1 = 0$$
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
$$a_{11} = \frac{1}{25}$$
$$a_{12} = 0$$
$$a_{13} = 0$$
$$a_{14} = 0$$
$$a_{22} = \frac{1}{16}$$
$$a_{23} = 0$$
$$a_{24} = 0$$
$$a_{33} = - \frac{1}{4}$$
$$a_{34} = 0$$
$$a_{44} = -1$$
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
$$I_{1} = - \frac{59}{400}$$
|1/25 0 | |1/16 0 | |1/25 0 |
I2 = | | + | | + | |
| 0 1/16| | 0 -1/4| | 0 -1/4|$$I_{3} = \left|\begin{matrix}\frac{1}{25} & 0 & 0\\0 & \frac{1}{16} & 0\\0 & 0 & - \frac{1}{4}\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}\frac{1}{25} & 0 & 0 & 0\\0 & \frac{1}{16} & 0 & 0\\0 & 0 & - \frac{1}{4} & 0\\0 & 0 & 0 & -1\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}\frac{1}{25} - \lambda & 0 & 0\\0 & \frac{1}{16} - \lambda & 0\\0 & 0 & - \lambda - \frac{1}{4}\end{matrix}\right|$$
|1/25 0 | |1/16 0 | |-1/4 0 |
K2 = | | + | | + | |
| 0 -1| | 0 -1| | 0 -1| |1/25 0 0 | |1/16 0 0 | |1/25 0 0 |
| | | | | |
K3 = | 0 1/16 0 | + | 0 -1/4 0 | + | 0 -1/4 0 |
| | | | | |
| 0 0 -1| | 0 0 -1| | 0 0 -1|$$I_{1} = - \frac{59}{400}$$
$$I_{2} = - \frac{37}{1600}$$
$$I_{3} = - \frac{1}{1600}$$
$$I_{4} = \frac{1}{1600}$$
$$I{\left(\lambda \right)} = - \lambda^{3} - \frac{59 \lambda^{2}}{400} + \frac{37 \lambda}{1600} - \frac{1}{1600}$$
$$K_{2} = \frac{59}{400}$$
$$K_{3} = \frac{37}{1600}$$
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
$$\lambda^{3} + \frac{59 \lambda^{2}}{400} - \frac{37 \lambda}{1600} + \frac{1}{1600} = 0$$
$$\lambda_{1} = - \frac{1}{4}$$
$$\lambda_{2} = \frac{1}{16}$$
$$\lambda_{3} = \frac{1}{25}$$
entonces la forma canónica de la ecuación será
$$\left(\tilde z^{2} \lambda_{3} + \left(\tilde x^{2} \lambda_{1} + \tilde y^{2} \lambda_{2}\right)\right) + \frac{I_{4}}{I_{3}} = 0$$
$$- \frac{\tilde x^{2}}{4} + \frac{\tilde y^{2}}{16} + \frac{\tilde z^{2}}{25} - 1 = 0$$
$$- \frac{\tilde x^{2}}{\left(\frac{2}{1}\right)^{2}} + \left(\frac{\tilde y^{2}}{\left(\frac{4}{1}\right)^{2}} + \frac{\tilde z^{2}}{\left(\frac{5}{1}\right)^{2}}\right) = 1$$
es la ecuación para el tipo hiperboloide unilateral
- está reducida a la forma canónica