Аналитическая геометрия. Часть III. Многомерные пространства. Гиперповерхности второго порядка. Шурыгин В.В. - 74 стр.

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(a
ij
) =
1 −1
−1 1
!
, (a
αβ
) =
1 −1 −5
−1 1 −3
−5 −3 25
.
I
2
= det(a
ij
) = 0
I
3
= det(a
αβ
) = −64
(1 − λ)
2
− 1 = 0 λ
1
= 2 λ
2
= 0
2x
2
+ 2
√
32y = 0
w
1
= {−1; 1} w
2
= {1; 1}
Ox
1
Ox
3
x
1
= t cos α x
2
= t sin α x
3
= a = const
M(x
1
; x
2
; x
3
) Ox
1
(x
2
)
2
+ (x
3
)
2
M
{cos α; sin α; 0} {x
1
; x
2
; x
3
−a}
(x
3
−a)
2
+(x
1
sin α−x
2
cos α)
2
= (x
2
)
2
+(x
3
)
2
2x
3
0
= −2ax
3
+ a
2
x
1
0
= x
1
sin α − x
2
cos α x
2
0
= x
2
2x
3
0
= (x
2
0
)
2
− (x
1
0
)
2
  ÈYÉY\]Y^ @5730½A 5//-½003-B5,,AC 1B5D3570),AC „-3: 0:*‡7 B0D
                                                              
                                   !                   1 −1 −5
                        1 −1                                  
            (aij ) =                   ,   (aαβ ) =  −1 1 −3  .
                       −1 1
                                                      −5 −3 25
:**: I = det(a ) = 0 < ’-¨7-:2 ¨7- 130B58 .535+-60)*/1-9- 70.5< 51
151 I3 = det(aαβ ) = −64  7- ¨7- F .535+-65< 53517*30/70)*/1-* 235B,*
        2          ij


,0* 0:**7 B0D (1 − λ)2 − 1 = 0 0:**7 1-3,0 λ1 = 2  λ2 = 0 < ’- „-3:26*
•I²‹ ,5C-D0: 15,-,0)*/1-* 235B,*,0* 2x2 + 2√32y = 0 < 68 ,5C-“D*,08
15,-,0)*/1-” /0/7*:A 1--3D0,57 ,2“,- /,5)565 .-B*3,27> -/0 1--3D0,57<
  *17-3A 965B,AC ,5.35B6*,0” 0:*‡7 B0D w1 = {−1; 1}  w2 = {1; 1} <
   ªœZœ«œ –ß^ -15E57>  )7- 9*-:*730)*/1-* :*/7- 7-)*1  35B,-2D56*,,AC
-7 DB2C /13*ˆ0B5‡ˆ0C/8 .38:AC  */7> 90.*3+-60)*/10” .535+-6-0D <
   ÈYÉY\]Y^ A+*3*: /0/7*:2 1--3D0,57 B 1-7-3-” .*3B58 .38:58 F -/>
Ox1  -+ˆ0” .*3.*,D012683 .38:AC F -/> Ox3  5 B7-358 .38:58 0:**7
235B,*,08 x1 = t cos α  x2 = t sin α  x3 = a = const < B5D357 35//7-8,08 -7
7-)10 M (x1; x2; x3) D- -/0 Ox1 35B*, (x2)2 + (x3)2  5 1B5D357 35//7-8,08
-7 M D- B7-3-” .38:-” 35B*, 1B5D3572 :-D268 B*17-3,-9- .3-0EB*D*,08
B*17-3-B {cos α; sin α; 0} 0 {x1; x2; x3 − a} < ’3035B,0B58 DB5 ¨70 )0/65 .-
62)0: 235B,*,0* (x3 −a)2 +(x1 sin α−x2 cos α)2 = (x2)2 +(x3)2 < 5„„0,,-”
/0/7*:* 1--3D0,57 2x3 = −2ax3 + a2  x1 = x1 sin α − x2 cos α  x2 = x2 ¨7-
                       0                        0                   0


235B,*,0* .30,0:5*7 B0D 2x3 = (x2 )2 − (x1 )2 <
                               0           0        0




                                           δ

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