简单分析一下，详情如图所示
两边对x求导得z'x-e^x+2y=0z'x=e^x-2y=e-4两边对y求导得z'y+2x=0z'y=-2所以切平面方程为-z'x(x-x0)-z'y(y-y0)+(z-z0)=0即(4-e)(x-1)+2(y-2)+z=0法线方程-(x-x0)/z'x=-(y-y0)/z'y=(z-z0)/1即(x-1)/(4-e)=(y-2)/2=z...
