Аналитическая геометрия. Часть III. Многомерные пространства. Гиперповерхности второго порядка. Шурыгин В.В. - 71 стр.

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1
9
1
9
ϕ
λ
1
(x
1
)
2
+ λ
2
(x
2
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2
+ a
33
= 0.
I
3
= λ
1
λ
2
a
33
= I
2
a
33
λ
1
(x
1
)
2
+ λ
2
(x
2
)
2
+
I
3
I
2
= 0.
ϕ
λ
1
6= 0 λ
2
= 0
λ
1
(x
1
)
2
+ 2a
23
x
2
= 0.
I
3
=
λ
1
0 0
0 0 a
23
0 a
23
0
, I
1
= λ
1
, a
23
=
r
I
3
I
1
.
λ
1
(x
1
)
2
+ 2
r
I
3
I
1
x
2
= 0.
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                                         I3
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