设f(x,y,z)=z-x^2-y^2+1那么f'(x)=-2xf'(y)=-2yf'(z)=1所以在点(2,1,4)处的法向量为(-4,-2,1)或(4,2,-1)法线方程为(x-2)/4=(y-1)/2=4-z切平面方程为4(x-2)+2(y-1)-z+4=0
