xz 2r2 交包 yfy 2515 (2y+2z) 画长设面垂足 (2y+2)=x2+(2y+2)2, 展开难包 25x2+4y2+z2+4yz-20y-10z=0 3.求曲线 对xOy平面的论影设面垂足 解:母线的垂向向量意(0,0,1).二包垂足组 xy2 x2+2y/2+ y2+y-1=0 平切y=-1影合题意,试正设面垂足圆 另交:将曲线垂足组的第应距垂足减去第二距垂足,二包2y2-y-1=0.这意母线法z轴平球的 设面,及且通过已过曲线,即圆卦求的论影设面 4.试说除面就垂足卦表向的曲面意设面: (1)(x+y)(y+2)=a2 (2)(x+y)(y+z)=x+2y+z; (3)y2+2y2+2=1-x2;(4)(x+y+2)2=(x-y-2) 解()试圆直线{2+2二0相正曲丽上它的垂向圆1:(-):1且点(:号,号)相正直线 上及平面x-y+2-=0法曲面(x+yy+2)=a2的交线圆 (x+y)(y+2)"    x = x 0 , y = y 0 + u, z = z 0 − 2u, y 0 = x 02 + z 02 , y 0 = 2z 0 , %=    x 0 = x, y 0 = 2 5 (2y + z), z 0 = 1 5 (2y + z), /03TU 2 5 (2y + z) = x 2 + 1 25 (2y + z) 2 , g= 25x 2 + 4y 2 + z 2 + 4yz − 20y − 10z = 0. 3. X ( x 2 + 2y 2 + z 2 = 1, x 2 + z 2 = y N xOy ;3,3TU. : TQQV (0, 0, 1). .=TU"    x = x 0 , y = y 0 , z = z 0 + u, x 02 + 2y 02 + z 02 = 1, x 02 + z 02 = y 0 , %=    x 0 = x, y 0 = y, 2y 02 + y 0 − 1 = 0, ;< y = −1 ,ra, ()3TU# y = 1 2 , Ã − √ 2 2 6 x 6 √ 2 2 ! . %: ITU"|BCTUYZ|.CTU, .= 2y 2 − y − 1 = 0. R z M;5 3, -$i:@:, t#FX,3. 4. (c34TUFPQ33: (1) (x + y)(y + z) = a 2 ; (2) (x + y)(y + z) = x + 2y + z; (3) y 2 + 2yz + z 2 = 1 − x 2 ; (4) (x + y + z) 2 = (x − y − z) 2 . : (1) (#
 ( x + y = a, y + z = a )3, ATQ# 1 : (−1) : 1. $& ³ a 2 , a 2 , a 2 ´ )
 . -;3 x − y + z − a 2 = 0 3 (x + y)(y + z) = a 2 %# ( (x + y)(y + z) = a 2 , x − y + z = a 2 . · 8 ·