先用T作旋转坐标变换,可得 :(2-) 即b=-√5 用X0作平移坐标变换 E 可得 0 0 E X 00 0 5 √5 050 即方程化简为5y2-2√5x”=0.其简化方程为y2 x".总的坐标变换公式为 √52√5 TXo 25√51y 这.抛物线的.点坐标TX0 10 新坐标系中x与y在旧坐标系中的方程分别 2x-y+2=0与2x+4y+1=0 A 5.化简下列二次曲线的方程并指出它.一什么曲线 (1)4x2-4xy+y2+4x-2y=0; 2y-3=0. 解()0矩阵A=(21),B=(2)因此4=m(4=5>0b==0h 21-1=0.为确.曲线 需要.步计算A的特征值方程A2-5=0的根,解得 A2=0=5、于征根0与5的单位持征向量分别(3)与(-2语),所以
{9 T x|} , -` µ T T 0 0 1 ¶ µ A B BT 3 ¶ µ T 0 0 1 ¶ = 0 0 − √ 5 0 5 −2 √ 5 − √ 5 −2 √ 5 3 . + b 0 1 = − √ 5, b 0 2 = −2 √ 5. K X0 = b 02 2 − λ2c 2b 0 1λ2 − b 0 2 λ2 = Ã − √ 5 10 2 √ 5 5 ! , 9 X0 x3~ µ X0 1 ¶ = µ E X0 0 1 ¶ µ X00 1 ¶ , -` µ E 0 XT 0 1 ¶ 0 0 − √ 5 0 5 −2 √ 5 − √ 5 −2 √ 5 −2 µ E X0 0 1 ¶ = 0 0 − √ 5 0 5 0 − √ 5 0 0 , +IJab 5y 002 − 2 √ 5x 00 = 0. 'baIJ y 002 = 2 √ 5 5 x 00 . µ X 1 ¶ = µ T T X0 0 1 ¶ µ X00 1 ¶ = √ 5 5 − 2 √ 5 5 − 9 10 2 √ 5 5 √ 5 5 1 5 0 0 1 x 00 y 00 1 . 0yzH 2 T X0 = ³ − 9 10 , 1 5 ´T . x 0 5 y 0 5M IJ!U2 2x − y + 2 = 0 2x + 4y + 1 = 0. 0 1 2 2 5 0 | / l . - - g ) (%$TPNKJHG F* & $ # uuur@ t t t t t t t t t p 8 PPPPPPPPPP !:F z / IX# x y x 0 y 0 O O0 4 5. abPEFGHIJ, c %2GH: (1) 4x 2 − 4xy + y 2 + 4x − 2y = 0; (2) x 2 − 2xy + y 2 + 2x − 2y − 3 = 0. : (1) :; A = µ 4 −2 −2 1 ¶ , B = µ 2 −1 ¶ , &C I1 = Tr(A) = 5 > 0, I2 = |A| = 0, I3 = ¯ ¯ ¯ ¯ ¯ ¯ 4 −2 2 −2 1 −1 2 −1 0 ¯ ¯ ¯ ¯ ¯ ¯ = 0. GH pq . A ijk2IJ λ 2 − 5λ = 0 l, W` λ1 = 0, λ2 = 5. rs[ijl 0 5 ABij!U2 µ √ 5 5 , 2 √ 5 5 ¶ µ − 2 √ 5 5 , √ 5 5 ¶ , () · 6 ·
2y5 且|=1 先用T{转坐标变换,可得 T0\/AB 01/(B 01 1=0.二取x=(-2)=()用x、平移坐标变换可得 ()(2)( 000 050 即方、化为52-1=0,即y=±52,这是标平行直线总的坐标变换公式为 2√5 0 (2)矩阵 因此I=Tr(A)=2>0,2=|41=0,I3 0.为确定曲线的类型、、坐标系与算.A的、征、是方 得=0=2、于、低0与2的单位征向量分别是(号)与(2,)所以 且 先用T1转坐标变换,可得 TT 0\/A B/T 0 0 01八(B-3八(01 即b=0,b 取x-(-%)() 用X0、平移坐标变换可得 E 0 E X 方化为2y2-4=0,即y"=±V2,这是标平行直线总的坐标变换公式为 01 6.求下、二次曲线的公近线 (1)6x2-xy-y2+3x+y-1=0 2)2ry-4x-2y+3=0
T = Ã √ 5 5 − 2 √ 5 5 2 √ 5 5 √ 5 5 ! , 7 |T| = 1. {9 T x|} , -` µ T T 0 0 1 ¶ µ A B BT 0 ¶ µ T 0 0 1 ¶ = 0 0 0 0 5 − √ 5 0 − √ 5 0 . + b 0 1 = 0, b 0 2 = − √ 5. K X0 = µ 0 − b 0 2 λ2 ¶ = µ √ 0 5 5 ¶ , 9 X0 x3~ , -` µ E 0 XT 0 1 ¶ 0 0 0 0 5 − √ 5 0 − √ 5 0 µ E X0 0 1 ¶ = 0 0 0 0 5 0 0 0 −1 , +IJab 5y 002 − 1 = 0, + y 00 = ± √ 5 5 , 02 r3OH. µ X 1 ¶ = µ T T X0 0 1 ¶ µ X00 1 ¶ = √ 5 5 − 2 √ 5 5 − 2 5 2 √ 5 5 √ 5 5 1 5 0 0 1 x 00 y 00 1 . (2) :; A = µ 1 −1 −1 1 ¶ , B = µ 1 −1 ¶ , &C I1 = Tr(A) = 2 > 0, I2 = |A| = 0, I3 = ¯ ¯ ¯ ¯ ¯ ¯ 1 −1 1 −1 1 −1 1 −1 −3 ¯ ¯ ¯ ¯ ¯ ¯ = 0. GH pq . A ijk2IJ λ 2 − 2λ = 0 l, W ` λ1 = 0, λ2 = 2. rs[ijl 0 2 ABij!U2 µ √ 2 2 , √ 2 2 ¶ µ − √ 2 2 , √ 2 2 ¶ , () T = Ã √ 2 2 − √ 2 √ 2 2 2 √ 2 2 ! , 7 |T| = 1. {9 T x|} , -` µ T T 0 0 1 ¶ µ A B BT −3 ¶ µ T 0 0 1 ¶ = 0 0 0 0 2 − √ 2 0 − √ 2 −3 . + b 0 1 = 0, b 0 2 = − √ 2. K X0 = µ 0 − b 0 2 λ2 ¶ = µ √ 0 2 2 ¶ , 9 X0 x3~ , -` µ E 0 XT 0 1 ¶ 0 0 0 0 2 − √ 2 0 − √ 2 −3 µ E X0 0 1 ¶ = 0 0 0 0 2 0 0 0 −4 , IJab 2y 002 − 4 = 0, + y 00 = ± √ 2, 02 r3OH. µ X 1 ¶ = µ T T X0 0 1 ¶ µ X00 1 ¶ = √ 2 2 − √ 2 2 − 1 √ 2 2 2 √ 2 2 1 2 0 0 1 x 00 y 00 1 . 6. $PEFGHH: (1) 6x 2 − xy − y 2 + 3x + y − 1 = 0; (2) 2xy − 4x − 2y + 3 = 0. · 7 ·