How do you find the equations for the tangent plane to the surface #x^2+4y^2+z^2=36# through #(2, -2, 4)#?

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Answer 1 · 2016-11-26T13:55:19

#Pi->4(x-2)-16(y+2)+8(z-4)=0#

The surface #f(x,y,z)=x^2+4y^2+z^2-36=0# has an associated gradient field given by

#grad f(x,y,z)=(f_x,f_y,f_z)=(2x,8y,2z)#

At point #p_0=(2,-2,4)# has the value

#vec n_0 =(4,-16,8)# so the tangent plane at #p_0# is given by

#Pi-> << p - p_0, vec n_0 >> = 0# where #p = (x,y,z)#

so

#Pi->4(x-2)-16(y+2)+8(z-4)=0#