Аналитическая геометрия. Часть III. Многомерные пространства. Гиперповерхности второго порядка. Шурыгин В.В. - 41 стр.

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a
n+1 j
= a
j n+1
A = (a
ij
) i, j = 1, . . . , n
e
A = (a
αβ
)
α, β = 1, . . . , n + 1
A = (a
ij
) x
i
x
j
e
A =
(a
αβ
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n = 2
A =
a
11
a
12
a
21
a
22
!
,
e
A =
a
11
a
12
a
13
a
21
a
22
a
23
a
31
a
32
a
33
A
e
A
Φ
F (x
k
)
Φ
A
n
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0
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0
i
}
A
n
x
i
= p
i
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i
0
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i
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F
0
(x
1
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, . . . , x
n
0
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F (x
k
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ij
x
i
x
j
+ 2a
i n+1
x
i
+ a
n+1 n+1
=
a
ij
(p
i
i
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x
i
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j
j
0
x
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+ b
j
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i n+1
(p
i
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0
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+ b
i
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n+1 n+1
=
(a
ij
p
i
i
0
p
j
j
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i
0
x
j
0
+ (a
ij
p
i
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x
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b
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b
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p
j
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x
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p
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)+
+(a
ij
b
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b
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x
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0
x
j
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+ 2a
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x
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(a
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a
i
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p
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p
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= F (b
1
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n
).
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       .30/27/7B2‡7 B/* 1-¨„„0½0*,7A < 5.30:*3  .30 n = 2 
                 ij
(aαβ )
                                                                                      
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                                       a11 a12                         e=             
                           A=                              ,           A  a21 a22 a23 
                                       a21 a22
                                                                           a31 a32 a33
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                                     xi = pii xi + bi .            0
                                                                        0 •²±‹
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           0         0                                                  
F 0 (x1 , . . . , xn ) = 0 90.*3.-B*3C,-/70 Φ B ,-B-” /0/7*:* 1--3D0,57< :**:

                                F (xk ) = aij xi xj + 2ai n+1 xi + an+1 n+1 =
                aij (pii0 xi + bi )(pjj 0 xj + bj ) + 2ai n+1 (pii0 xi + bi ) + an+1 n+1 =
                            0                 0                                        0



                  (aij pii0 pjj 0 )xi xj + (aij pii0 xi bj + aij bi pjj 0 xj + 2ai n+1 pii0 xi )+
                                   0      0                    0                           0         0


                                                                        0    0                   0       0
+(aij bi bj + 2ai n+1 bi + an+1 n+1 ) = ai0 j 0 xi xj + 2ai0 n+10 xi + an+10 n+10 = F (xk ).
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                                                                    ij




 a = a pi pj , a     = (a bj +a    )pi , a        = F (b1 , . . . , bn ). •
                                                                           ²²‹
  i0 j 0       ij i0 j 0        i0 n+10           ij           i n+1        i0       n+10 n+10


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