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1、1?IC?:C? 1?IC? 1.?IXO;1,2?O; 1, 2k?:. 3?IXO;1,2 e,#?IX?: 1= 2 2 2 2 , 2= 2 2 2 2 . (1)?IC?“; (2)?IX?1= 1 0 ! , 2= 0 1 ! 3#?IXe? I?; (3) fi ? v3O;1,2? 1 1 ! ,3#?IXO; 1, 2 e?. ): (1)?( 1, 2) = (1,2)T, ?T = 2 2 2 2 2 2 2 2 , ?I C?“? x y ! = T x y ! = 2 2 2 2 2 2 2 2 x y ! . (2)d(1)?: (1,2) = ( 1, 2)T 1
2、, ?T1= 2 2 2 2 2 2 2 2 , 106 1?IC?:C? 1= 2 2 1 2 2 2, 2 = 2 2 1 + 2 2 2, =: 1= 2 2 2 2 , 2 = 2 2 2 2 . (3)l x y ! = T x y ! ? x y ! = T1 x y ! .y3 v = 12, x y ! = 2 2 2 2 2 2 2 2 1 1 ! = 0 2 ! , ? v3#?IXe?. 2.3?IXO;1,2, fi ?#?IXO; 1, 2? ?:O?I?(3,2),:M(5,3)3#?IX?x?,?:M?#?I x 0.?/“?lO;1,2?O; 1, 2?IC?“
3、. ):?X0= 3 2 ! ,?dK? OM = 5 3 ! 3 2 ! = 2 1 ! , ? OM? 25 5 5 5 , =cos = 25 5 , sin = 5 5 .T = 25 5 5 5 5 5 25 5 . dC?“? x y 1 = 25 5 5 5 3 5 5 25 5 2 001 x y 1 . 3.?g?C3?IXO;1,2?: x2 xy + y2+ 2x 4y + 1 = 0. 1 ?IC? 107 (1)?#?IXO; 1, 2, O3?IXe?I?(0,2),? k 1= 2 2 1+ 2 2 2 2= 2 2 1+ 2 2 2, ?/“?IC?“; (2)
4、?C3#?IXe?. ): (1)?K?, X0= 0 2 ! , T = 2 2 2 2 2 2 2 2 , ?1?“|T| = 1. ?IC?“?: x y 1 = 2 2 2 2 0 2 2 2 2 2 001 x y 1 . (2) C? x2 6 + y2 2 = 1. 4.?k?IXO;1,2,e#?IXO; 1, 2?v: x ?y3?IX?Ox 2y + 2 = 0?2x + y + 4 = 0. (1)l?IX?#?IX?C?“; (2)?x y + 2 = 03#?IX?; (3)?3x+ y+ 1 = 03?IX?. ): (1)O:?I(x0,y0)?| x 2y +
5、2 = 0 2x + y + 4 = 0 ?),= X0= x0 y0 ! = 2 0 ! . w,x 2y + 2 = 0?X?2 : 1, 2x + y + 4 = 0?X ?1 : 2. 1 = 25 5 1+ 5 5 2, 2 = 5 5 1+ 25 5 2,=( 1, 2) = (1,2) 25 5 5 5 5 5 25 5 . du ? ? ? ? ? ? ? 25 5 5 5 5 5 25 5 ? ? ? ? ? ? ? = 1, 1, 2 ?mX. ?IC?“? 108 1?IC?:C? x y 1 = 25 5 5 5 2 5 5 25 5 0 001 x y 1 . (2)
6、 x y + 2 = 03#?IX?: 25 5 x 5 5 y 2 ! 5 5 x+ 25 5 y ! + 2 = 0, =x 3y= 0. (3)d(1)?IC?“? x y 1 = 25 5 5 5 45 5 5 5 25 5 2 5 5 001 x y 1 , ?3x+ y+ 1 = 03?IXe?: 3 25 5 x + 5 5 y + 45 5 ! + 5 5 x + 25 5 y 25 5 ! + 1 = 0, =5x + 5y + 10 + 5 = 0. 2?g?z 1.z?g? 5x2+ 4xy + 2y2 24x 12y + 12 = 0, x?/9#?I. ):?A =
7、52 22 ! , B = 12 6 ! ,dI1= Tr(A) = 7 0, I2= |A| = 6 0, I3= ? ? ? ? ? ? ? 5212 226 12612 ? ? ? ? ? ? ? = 108 0.d?. A ?A?2 7 + 6 = 0?,)?1= 6, 2= 1.?z? ?6x2+ y2 108 6 = 0,= x2 3 + y2 18 = 1. 2 ?g?z 109 ?x?/,I?IC?“.AuA?6?1?A ?O 25 5 , 5 5 ! ? 5 5 , 25 5 ! ,T = 25 5 5 5 5 5 25 5 , ?|T| = 1.2?%(=#?IX?:) O
8、(x0,y0),)?5?| 5x0+ 2y0 12 = 0 2x0+ 2y0 6 = 0 ?X0= x0 y0 ! = 2 1 ! .d?IC?“ x y 1 = 25 5 5 5 2 5 5 25 5 1 001 x y 1 . ?#?IXx?y3?IX?Ox 2y = 0 (= y= 0)?2x + y 5 = 0 (=x= 0). x y x y O O 1 1 K 2.z?g? x2 4xy 2y2+ 10 x + 4y = 0, x?/9#?I. 110 1?IC?:C? ):?A = 12 22 ! , B = 5 2 ! ,dI1= Tr(A) = 1 0, I2= |A| =
9、6 0.d?V?. A?A ?2+ 6 = 0?,)?1= 2, 2= 3.?z? 2x2 3y2 1 = 0,= x2 1 2 y2 1 3 = 1. ?x?/,I?IC?“.AuA?2?3? A?O 25 5 , 5 5 ! ? 5 5 , 25 5 ! ,T = 25 5 5 5 5 5 25 5 , ?|T| = 1.2?%(=#?IX?:) O(x0,y0),)?5?| x0 2y0+ 5 = 0 2x0 2y0+ 2 = 0 ?X0= x0 y0 ! = 1 2 ! .d?IC?“ x y 1 = 25 5 5 5 1 5 5 25 5 2 001 x y 1 . ?#?IXx?y
10、3?IX?Ox + 2y 3 = 0? 2x y + 4 = 0. 2 ?g?z 111 x y x y O O 1 2 K 3.z?g? x2 3xy + y2+ 10 x 10y + 21 = 0, ?/9#?I. ):?A = 13 2 3 2 1 , B = 5 5 ! ,dI1= Tr(A) = 2 0, I2= |A| = 5 4 0, I3= ? ? ? ? ? ? ? ? ? 13 2 5 3 2 15 5521 ? ? ? ? ? ? ? ? ? = 5 4 0, I2= |A| = 0, I3= ? ? ? ? ? ? ? 423 214 343 ? ? ? ? ? ? ?
11、 = 25 0, I2= |A| = 0, I3= ? ? ? ? ? ? ? 422 211 210 ? ? ? ? ? ? ? = 0.?(?a.I? O?. A?A?2 5 = 0?,)?1= 0, 2= 5.Au A?0?5?A?O 5 5 , 25 5 ! ? 2 5 5 , 5 5 ! , T = 5 5 2 5 5 25 5 5 5 , ?|T| = 1. kT?=?IC?,? TT0 01 ! AB BT0 ! T0 01 ! = 000 055 050 . =b1= 0, b2= 5.?X0= 0 b 2 2 = 0 5 5 , X0?IC 2 ?g?z 115 ?,? E0
12、 XT 0 1 ! 000 055 050 EX0 01 ! = 000 050 001 , =?z?5y2 1 = 0,=y= 5 5 ,?1?.o?IC? “? X 1 ! = TTX0 01 ! X 1 ! = 5 5 2 5 5 2 5 25 5 5 5 1 5 001 x y 1 . (2)?A = 11 11 ! , B = 1 1 ! ,dI1= Tr(A) = 2 0, I2= |A| = 0, I3= ? ? ? ? ? ? ? 111 111 113 ? ? ? ? ? ? ? = 0.?(?a.I? O?. A?A?2 2 = 0?,)?1= 0, 2= 2.A uA?0
13、?2?A?O 2 2 , 2 2 ! ? 2 2 , 2 2 ! , T = 2 2 2 2 2 2 2 2 , ?|T| = 1. kT?=?IC?,? TT0 01 ! AB BT3 ! T0 01 ! = 000 022 023 . =b1= 0, b2= 2.?X0= 0 b2 2 = 0 2 2 , X0?IC ?,? E0 XT 0 1 ! 000 022 023 EX0 01 ! = 000 020 004 , 116 1?IC?:C? ?z?2y2 4 = 0,=y= 2,?1?.o?IC? “? X 1 ! = TTX0 01 ! X 1 ! = 2 2 2 2 1 2 2
14、2 2 2 1 2 001 x y 1 . 6.e?g?C?: (1) 6x2 xy y2+ 3x + y 1 = 0; (2) 2xy 4x 2y + 3 = 0. ): (1)?A = 61 2 1 2 1 , B = 3 2 1 2 , dI2= |A| = 25 4 0.d?V?.?%(x0,y0)? v?5?| 6x0 1 2y0 + 3 2 = 0 1 2x0 y0+ 1 2 = 0 ?X0= x0 y0 ! = 1 5 3 5 . ?C? 6 ? x + 1 5 ?2 ? x + 1 5 ? y 3 5 ? ? y 3 5 ?2 = 0. ?“?)? ? 3 ? x + 1 5 ? + ? y 3 5 ? 2 ? x + 1 5 ? ? y 3 5 ? = 0, ?C?3x + y = 0?2x y + 1 = 0. (2)?A =
