先用T作旋转坐标变换，可得 969-(日 即以=-V5,防=-2V5.取 -(）() 用X0作平移坐标变换 ()-(6)() 可得 低6-( 即方程化简为52-2V5”=0.其简化方程为？=5.总的坐标变换公式为 --摩￥间 这条抛物线的顶点坐标是TX。=（一品)'.新坐标系中口轴与y轴在旧坐标系中的方程分别是 2x-y+2=0与2x+4g+1=0. 第4题图 5.化简下列二次曲线的方程并指出它们是什么曲线： (1)4r2-4红y+32+4z-2y=0: @2-2++2-223=0 解国矩阵A=(3)B=()因此4==5>0五=W=06- 4-221 -2】-1=0为确定曲线的类型需要进一步计算A的特征值是方程2一5=0的根解得 2-10 1=0,如=5.对应于特征根0与5的单位特征向量分别是(，2）与(-25，)，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 ·
r-()m 先用T作旋转坐标变换，可得 G9台g》- 即丝=0,6=-6取X。=气_是)=（经)）用X作平移坐标变换可得 (号Gg)G /000 即方程化简为52-1=0，即”=士5，这是一对平行直线.总的坐标变换公式为 ()-(6))- g)矩阵A=(已1)B=（已)因此h=T)=2>0h=M=0,6- 1】-=心为确定由浅的类室需要进-步计算A的特征值起方程P一2以=0的根解 得1=0，如=2对应于特征根0与2的单位特征向量分别是（要，号）与（-要，号）所以 -()m 先用T作旋转坐标变换，可得 9()6)-(68 即以=06=-反取x=(_是)=（是）用0作平移坐标变换可得 0-2-3/ 100-4 方程化简为2-4-0,即”-±V2,这是一对平行直线.总的坐标变换公式为 -g-)间 6.求下列二次曲线的渐近线： (1)6x2-xy-y2+3红+y-1=0 (22xy-4r-2y+3=0. 7
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 ·