第三章第二节二次曲线类型

x2 y22bx2byc Fxyax22axyay22bx2by 1.1设Ⅱ[O;e',e'是由Ⅰ y cosy'xy cosy' ax22axyay2 x cos2a12(cosxsiny)(sinxcosax22axyay2

a11(sinxcos

siny)cosy) 2 12

2 (a11 a12a22 )[(a22a11sin22a11cos2]xy(a22a11)sin22a11cos20cot2a11 x2 y22bx2byc ax2 2? 2b1x2b2yc2b1(cosxsinxcosxsinyy 令cot2a11a22 a22

b1b1cosb2sin

sinb2 '2 '22b'x'2b'y'c 例1作转轴消去二次方程中的交叉项5x24xy2y224x12y180解：选取角满足：cot2a11 5 2tan23tan2解 tan2

(tan2舍去)， sin 1 cos 52a11a11a12tan5212a22a22a12tan2212 1 12

b

6

2 c5x24xy2y224x12y180变换公式为：xy)tCx 与a皆不为 (x'

1)2

(y'

2

c

a','x' a','

x2 x2y2c y

b'

b' c(1

2)'''2a22''22b''1' 12y'c 方程化为 2 2c a2 1x2a2 1 线

直 直

1

a2 a2 情形2.a'与

x'22bx2b

c

b若2 0b若2

x22

x'

' x22b'y' c yy'

'b'

若b0，方程为 a(x a2ab'2bcc

..

x

x'

a'1

y

xx211

2c2当d当d

ozozx2y2a2 a2 c2x5x24xy2y224x12y18cot23,tan1

)2y2126x2y212 短轴在x上，半短轴长为

x 5

yy

55x 155y yxoyo(2,1ox2而tan1ox2例2x23xyy210x10y21cot2a11

＝110

cos

2 tana11a11a12tan131 22a22a12tan131 a

a2 c y y21y x25(y22)22 xx,yy x25y22x22

5 y 5 x 2 y

x x

y yy8x26xy26x12y134x28xy4y26x10y13x26xy4y26x4y12作业作业： x2 y22bx2byc Fxyax22axyay22bx2by F(x,y)(

a12x2bx2by y 22

见P b1x1(x,y,1) by1

1

bbb1c20 a0

A

c则A可分块写成A c 再令

x,yF(x,y)(t

c1

,1)

c

1令C sin,

x', cos y y' 00x y

sin 0x' 0 '111

1y

0' 1 11 (t,1) c1 0 0't,1)

c

1 CA CtCt ('t t 1方程的二次项系数及转角有关；新方数及转角有关；但常数项不变。由CtAC )sincos (cos2sin2 .当 CtA Ct'.

t1('t,1) t

10新方程交 x'2a'y'22bx'2by'c b

b

b2sinb 'b2

那么a, ?a,a是CtAC对角元

是正交矩阵，所以CtC 0 0 C1AC ACC 0 0

a'

a sin sin 0

a12 cos cos

22 22 cos sina sin cos 假如cos0，则由上式得a12sin ，因此得sin 于是cos

a' ' '