03_autovalores

1 0
0 0 1
\uf8f9\uf8fa\uf8fb .
Assim, definindo
\uf8ee\uf8ef\uf8f0 x\u2032y\u2032
z\u2032
\uf8f9\uf8fa\uf8fb =
\uf8ee\uf8ef\uf8f0 \u22121/
\u221a
2 1/
\u221a
2 0
\u22121/
\u221a
6 \u22121/
\u221a
6 2/
\u221a
6
1/
\u221a
3 1/
\u221a
3 1/
\u221a
3
\uf8f9\uf8fa\uf8fb
\uf8ee\uf8ef\uf8f0 xy
z
\uf8f9\uf8fa\uf8fb ,
7
podemos escrever a equc¸a\u2dco da qua´drica como
[
x\u2032 y\u2032 z\u2032
]\uf8ee\uf8ef\uf8f0 2 0 00 2 0
0 0 8
\uf8f9\uf8fa\uf8fb
\uf8ee\uf8ef\uf8f0 x\u2032y\u2032
z\u2032
\uf8f9\uf8fa\uf8fb+ [ \u22121 0 0 ]
\uf8ee\uf8ef\uf8f0 \u22121/
\u221a
2 \u22121/
\u221a
6 1/
\u221a
3
1/
\u221a
2 \u22121/
\u221a
6 1/
\u221a
3
0 2/
\u221a
6 1/
\u221a
3
\uf8f9\uf8fa\uf8fb
\uf8ee\uf8ef\uf8f0 x\u2032y\u2032
z\u2032
\uf8f9\uf8fa\uf8fb = 3,
que equivale a
2x\u20322 + 2y\u20322 + 8z\u20322 +
\u221a
2
2
x\u2032 = 3.
Completando quadrado em x\u2032 chegamos a equac¸a\u2dco
2
(
x\u2032 +
\u221a
2/8
)2
+ 2y\u20322 + 8z\u20322 =
49
16
,
que pode ser simplificada(
x\u2032 +
\u221a
2/8
)2
(7
\u221a
2/8)2
+
y\u20322
(7
\u221a
2/8)2
+
z\u20322
(7
\u221a
2/16)2
= 1,
No sistema de coordenadas x\u2032y\u2032z\u2032 reconhecemos que a equac¸a\u2dco da qua´drica e´ um
elipso´ide, com semi-eixos a = b = 7
\u221a
2/8 e c = 7
\u221a
2/16 e centrada em x\u2032 = \u2212
\u221a
2/8,
y\u2032 = 0 e z\u2032 = 0.
Exerc´\u131cio 4.1 Identifique e esboce (explicando o desenho) as seguintes qua´dricas:
\u2022 2xy + 2xz + 2yz \u2212 6x\u2212 6y \u2212 4z = \u22129
\u2022 7x2 + 7y2 + 10z2 \u2212 2xy \u2212 4xz + 4yz \u2212 12x+ 12y + 60z
\u2022 2xy \u2212 6x+ 10y + z \u2212 31 = 0
\u2022 2x2 + 2y2 + 5z2 \u2212 4xy \u2212 2xz + 2yz + 10x\u2212 26y \u2212 2z = 0
8