设切点(x0,y0,z0)Fx=2xFy=4yFz=2z法向量(2x0,4y0,2z0)即:切平面2x0(x-x0)+4y0(y-y0)+2z0(z-z0)=0平行于已知平面则2x0/1=4y0/-1=2z0/22x0=-4y0=z0代入椭球面得出切点(-√2/11,√1/22,-√8/11)或(√2/11,-√1/22,√8/11)切平面-√8/11(x+√2/11)+√8/11(y-√1/22)-√32/11(z+√8/11)=0或√8/11(x-√2/11)-√8/11(y+√1/22)+√32/11(z-√8/11)=0解得切平面-(x+√2/11)+(y-√1/22)-2(z+√8/11)=0或(x-√2/11)-(y+√1/22)+2(z-√8/11)=0化简-x+y-2z-√(11/2)=0或x-y+2z-√(11/2)=0