上正方程减去方程x+y+之=1的2倍后包到 x2+2+22-2x-2y-2=0, 说明正球面与平面x+y+z=1的交线就是S2.试正S与S2相同应距球面上 *7.证明：过圆： 5-r2+y2+z2+2ur+2y+2z+d=0,(u2+2+m2-d>0), E=A+Bu+Cz+D=0. (A2+B2+C2≠0) 的球面族方程二表示圆：S+2八E=0(A圆参数)， 证明：对切参数入若方程5+迈二0确实表示应距球面，则圆{。应定相正球面中 对任意应距过圆{的球看 E=0 x2+2+2+2pr+2gy+2tz+r=0, 已过圆应相已距球的交线 x2+2+22+2+2y+2wz+d=0, x2+w2+x2+2pz+2则+2Hz+r=0 上平第4题过，正圆应相平面 p-r+(g-+-z+T=0 上.但E-0也过正圆，试正已距平面重合，即存相实数入使包 p-W-M.q-0=3B,I-w-XC.d=D. 解长，,t,,代入球面方程即包 x2+2+z2+2ur+2py+2z+d+2A(4z+Bg+Cz+D)=0. 到直6-3 1.求面就旋转曲面的方程： ()直线号-头=.1绕直线x==旋转 1 (2)直线工 专绕z轴旋转 「=2px, (3)抛物线 z=0 绕它的准线旋转 x2= (4)曲线 +:=0绕直线子=号=子旋转 解()显然端点0相旋转轴上，且轴的方向向量是=(1,1,1).参照(3.1)，二限包到方程组 (-)+(g-)+(2-)=0, 2+2+22=x2+2+22, (号=生=1 相方程组标消去参数x,,x后二包 2++2-1=e+y+2-1 试正卦求旋转曲面的方程圆 2(r2+y2+23)-5(xy+x2+y2)+5(x+y+2-7=0, 4. )TUYZTU x + y + z = 1  2 ng=￾ x 2 + y 2 + z 2 − 2x − 2y − 2 = 0, c)53;3 x + y + z = 1 %4 S2. () S1  S2 EBC53. ∗7. : :#: ( S = x 2 + y 2 + z 2 + 2ux + 2vy + 2wz + d = 0, (u 2 + v 2 + w 2 − d > 0), E = Ax + By + Cz + D = 0, (A2 + B2 + C 2 6= 0) 53oTU.PQ#: S + 2λE = 0 (λ #b ). : N<b λ, cTU S + 2λE = 0 PpPQBC53, # ( S = 0, E = 0 Bm)53. NBC:# ( S = 0, E = 0 53 x 2 + y 2 + z 2 + 2px + 2qy + 2tz + r = 0, @:#B@C5% ( x 2 + y 2 + z 2 + 2ux + 2vy + 2wz + d = 0, x 2 + y 2 + z 2 + 2px + 2qy + 2tz + r = 0 . ;| 4 a:, )#B;3 (p − u)x + (q − v)y + (t − w)z + r − d 2 = 0 . q E = 0 :)#, ()@C;3!r, tp λ = p − u = λA, q − v = λB, t − w = λC, r − d 2 = λD. %0 p, q, t, r, JK53TUt= x 2 + y 2 + z 2 + 2ux + 2vy + 2wz + d + 2λ(Ax + By + Cz + D) = 0. ￾
6–3 1. X34s3TU: (1)
 x 2 = y 1 = z − 1 0 t
 x = y = z s; (2)
 x − 1 1 = y −3 = z 3 t z Ms; (3) uv ( y 2 = 2px, z = 0 tAws; (4)  ( x 2 = y, x + z = 0 t
 x 1 = y 2 = z 1 s. : (1) xy7& O sM, $MTQQV ξ = (1, 1, 1). bx (3.1), .G=￾TU"    (x − x 0 ) + (y − y 0 ) + (z − z 0 ) = 0, x 2 + y 2 + z 2 = x 02 + y 02 + z 02 , x 0 2 = y 0 1 = z 0 − 1 0 , TU" yZb x 0 , y0 , z0 g.= x 2 + y 2 + z 2 − 1 = 5 9 (x + y + z − 1)2 , ()FXs3TU# 2(x 2 + y 2 + z 2 ) − 5(xy + xz + yz) + 5(x + y + z) − 7 = 0. · 4 ·