高等代数与解析几何第九章答案

0
2 b p
2
^
p



\ t p 5

X

p
[
9
Z p P W V U SR E p H
PNM
t
1
·3·
2.
abEFGHIJ
. −2 , B = −2
cd%e<)f 1 : :; A = −2
1 −2 5 −2 5 −2 2 = 6 > 0. 2 0
2
5
x2 − 4xy − 2y 2 + 10x + 4y = 0, 5 , 2
p
m
y
7 p m H
x
( 0
x
p
o
0
0
w
p
o
0
0
v t
p
p 0
0
v
p
0
q
0
t
p
0
r
0
t
p
0
s
0
t
s
p 0
t
0
r
0
0
q 0
0
O
p
t
p
u
q 0
p
v
0
t
o 0
p
w
0
p x 0 uuuuuuu @P
o
! F
0
0
0
0
O
:
n
x
p
y
l
p
y
0
t
l
p
n
/
. k p
n
n
i
p /

h
p
/
0 t f p

d p
O
:
t s
x x
j
V
t
t
t
3
4.
abEFGHIJ
4x2 − 4xy + y 2 + 6x − 8y + 3 = 0, −2 , B = 1
cd%e<)f 4 : :; A = −2
4 −2 3 −2 1 −4 3 −4 = −25 < 0. 3
5.
3 , −4
&C I
1
= Tr(A) = 5 > 0, I2 = |A| = 0, I3 =
1.
2 5 5 4 5 x+ y+ 5 5 5
IJ : √ √
+
√ √ 5 2 5 2 5 − x+ y− 5 5 5
√
+ 1 = 0,
9–2
abEFGHIJ
5x2 + 4xy + 2y 2 − 24x − 12y + 12 = 0, −12 , −6
cd%e<)f 5. 5 2 : :; A = , B = 2 2
IJ x6 + y2 = 1. 4. D 34 [O; η , η ], Q [O ; η , η ] RS: x 5T y 5M IJ!U2 x − 2y + 2 = 0 T 2x + y + 4 = 0. (1) $,M = ; (2) $ H x − y + 2 = 0 IJ; (3) $ H 3x + y + 1 = 0 M IJ. x − 2y + 2 = 0 −2 W, + X = x = . : (1) & O (x , y ) 2IJV y 0 2x + y + 4 = 0 √ √ 2 5 5 XY x−2y+2 = 0 I Z 2 : 1, 2x+y+4 Z −1 : √ 2. () η √= 0 I √ √= 5 η + 5 η , 2 5 − 5 2 5 − 5 √ √ 5 5 5 5 √ √ √ η = − 5 η + 2 5 η , + (η , η ) = (η , η ) . *[ √ = 1, η , η \] 5 5 5 2 5 5 2 5 5 5 5 5 ^_ . ()
1 2 1 2 1 2 1 2 0
+
cos θ = 2 5 , sin θ = 5 . 5 5
√
√
()
√ √ 2 5 − 5 5 5 √ √ T = . 5 2 5 5 5√ √ 2 5 − 5 3 x x 5 5 √ √ y . y = 5 5 2 2 5 5 1 1 0 0 1
&C I
2
1
= Tr(A) = −1 < 0, I2 = |A| = −6 < 0, I3 =
2
CGH2wGH. A ijk2IJ λ
2 2
+λ−6 = 0
l, W` λ
1
= 2, λ2 = −3.
@bamIJ 2x − 3y − 1 = 0, + x1 − y1 = 1. 2 3 d'noe<, pq$ √ . rs[ijl 2 −3 ABij!U2 √ 2 5 5 √ √ √ √ 2 5 ,− 5 5 , 2 5 , () T = 5 5 √ √ , 7 |T | = 1. t$GH u (+ 5 5 5 5 5 2 5 − 5 5 ) O (x , y ), WHvIJV
, B =
&C I
1
= Tr(A) = 2 > 0, I2 = |A| = − 5 < 0, 4
2
5 −5 = − 4 < 0. 21
CGH2wGH. A ijk2IJ λ
− 2λ − 5 = 0 4
l, W`
λ1 = 5 , λ2 = − 1 . 2 2
@bamIJ 5 x 2
√ √ 2, 2 2 2
2
− 1 y 2 + 1 = 0, 2
+ y2
2
2 − x 2 = 1. 5
d'noe<, pq$
√
. rs[ijl
2 2 √ − 2 2 √ √ 2 2 √ 2 2 ,
5 2
−1 2
ABij!U2
) O (x , y ), WHvIJV
0 0
√ 2 ,− 2 2 2
,
() T =
1 2 1 2 0 0 0 0 0 1 1 2 2 1 2 1 2 1 2 1 2
(2) x − y + 2 = 0

+ x − 3y = 0. (3) * (1)
√ √ 2 5 − 5 −2 x x 5 5 √ √ y . y = 5 2 5 0 5 5 1 1 0 0 1 : √ √ √ √ 2 5 5 5 2 5 x − y −2 − x + y + 2 = 0, 5 5 5 5
0 0
x0 − 2y0 + 5 = 0 −2x0 − 2y0 + 2 = 0
`X
0
=
x0 y0
=
−1 . 2
&C
$
x
5
y
5M
√ 2 5 x 5 √ y = − 5 5 1 0
2
IJ!U2 x + 2y − 3 = 0
y
z 8
5 5 √ 2 5 5 0
L M
89:;< (2) $GH C IJ. √ 2 0 2 √ : (1) N>D, X = ,T = 2 2
0
7OP |T | = 1. ()
x y . 2 1 1 0
:
x y = 1 (2) C
2 2
2
5 2 −12 2 2 −6 −12 −6 = −108 < 0. 12 λ2 = 1.
2
&C I
1
= Tr(A) = 7 > 0, I2 = |A| = 6 > 0, I3 =
2
CGH2gh. A ijk2IJ λ
2 2 2
− 7λ + 6 = 0
l, W` λ
1
= 6,
y @bamIJ 6x + y − 108 = 0, + x + = 1. 6 3 18 d'noe<, pq$ . rs[ijl 6 1 ABij!U2 √ √ 2 5 − 5 √ √ √ √ 2 5, 5 5 5 √ √ , 7 |T | = 1. t$GH u (+ − 5 , 2 5 , () T = 5 5 5 5 2 5 5 5 5 ) O (x , y ), WHvIJV
IJ
-)`=

√ 2 5 x 5 √ y = − 5 5 1 0
5 5 √ 2 5 5 0
√
√ 4 5 Fra Baidu bibliotek √ −2 5 5 1

x y , 1
·2·
@ 3x + y + 1 = 0 M √
3
+ 5x + 5y + 10 + √5 = 0.
√
x y . 1 1 1 2

= 0)
2x + y − 5 = 0 (
+
y
w h ,
z
p
y
/ MPQRJ p L H p
-
p
p
p p

C

<
p
Z
8 \ t p
_

p 4
a
b

p 2
1
b
p f

0 t p g
0
/
o
p
i
|
p
k
n
p /
k
m
z t
&C
5
·1·
3.
DEFGH C K
[O; η1 , η2 ]
IJ2: (0, 2), 7
x2 − xy + y 2 + 2x − 4y + 1 = 0. (1) [O ; η1 , η2 ], O √ √ η1 = 2 η1 + 2 η2 2√ 2√ η2 = − 2 η1 + 2 η2 , 2 2 ; √ − 2 √2 , 2 2 √ √ 2 − 2 2 √ √2 2 2 2 2 0 0
0 0
5x0 + 2y0 − 12 = 0 2x0 + 2y0 − 6 = 0
`X
0
=
x0 y0
=
2 . 1
&C
$
x = 0).
x
5
y
5M
√ 2 5 x 5 √ y = 5 5 1 0
2
−
IJ!U2 x − 2y = 0 (+ y
V P T [
5 5 √ 2 5 5 0
√
−1
x y . 2 1 1

2x − y + 4 = 0.
y
/

r



s p

r

s

s

s t

t

u

v

t l
m
P
/

o
P
P


P 4
8
7

P

7
6 t P 5 J P I
E
H
2
5
O

H P E P
P
8 P

P

P
/ P m t
P .
k P
v
X P # I
u
t
t s
!
F
uuuu@


O
t t
s
:
x x
s
r

s
r
r
r
r
r
r
r
t r
r
r
i
,
t
2
3.
abEFGHIJ
−3 2 1
cx%e<)f 1 : :; A = −3
1 I3 = − 3 2 5 ·4· −3 2 1 −5 5 2
5.
x2 − 3xy + y 2 + 10x − 10y + 21 = 0, 5 , −5
y
0 / t z 6
2 2 √ 2 2 0
−2
x y . 2 1 1 x − y + 4 = 0.
t
t
*
y

q
s
4
q

3 s
t


q
.
, 6
s
&
8
8 #S : PON
q
t s
q
s
q
O
q
t s
s
OPRU : 9 & *
q -
/ t s

1
q
s

4
!F
uuuuu q
& (η , η ) = (η , η )T , '
T =
2 2 √ 2 2 √
, $% √ √
!; [O; η , η ] !.
1 2

()

* #
'
() η ()
√
1
=
√ 2η − 2η , 2 1 2 2
+
,
-.#
/
012 ! 2. 34 [O; η , η ] , "# [O ; η , η ] O (3, 2), M (5, 3) x 56, 7 M x > 0. 89:;<, [O; η , η ] = [O ; η , η ] . √ 2 5 − − → − − → 3 5 3 2 5 √ : & X = , 7*>?# O M = , − = , @ O M AB2 5 2 3 2 1
9–1
:
1.
[O; η1 , η2 ]
[O; η1 , η2 ] 2 2 √ 2 2 √
.
, η2 = 0 1 √ − 2 √2 2 2 .
[O; η1 , η2 ]
,

η1 = (1)
; (2) η = 1 0 → (3) "# − v [O; η , η ] !
7 |T | = 1. t$GH u (+
`X
0
=
x0 y0
=
−2 . 2
&C
$
x
5
y
5M
√ 2 x 2 √ y = − 2 2 1 0
2
x0 − 3 y0 + 5 = 0 2 3 − x0 + y0 − 5 = 0 2 √
IJ!U2 x + y = 0
1 1 2
, η2 = 1 −1
− 2 √2 , 1 2 1 2 2 2√ 2 − 2 x x x 2 √2 =T = √ . 2 2 y y y 2 2 √ √ 2 2 −1 −1 2 2 √ √ T = , (2) (1) : (η1 , η2 ) = (η1 , η2 )T , 2 − 2 √ √ 2 2 2 2 √ √ 2 2 2 2 √ η2 = η + η , : η1 = , η2 = √ . 2 1 2 2 2 − 2 2 2 x x x x − → −1 (3) =T =T . v = η1 − η2 , y y y y √ √ 2 2 x 1 0 2 2 √ √ √ = = , 2 y −1 − 2 − 2 2 2 − → v . : (1)