Write an equation for the plane tangent at the point $P(2,1,1)$ to the surface with the equation $x^3+y^3+z^3)=5xyz$

So I have implicitly differentiated this wrt x and y. And obtained $z_x=1$ and $z_y=-1$ at the given point. I'm just a little confused how to use that to get the plane equation!

Question

Write an equation for the plane tangent at the point $P(2,1,1)$ to the surface with the equation $x^3+y^3+z^3)=5xyz$

So I have implicitly differentiated this wrt x and y. And obtained $z_x=1$ and $z_y=-1$ at the given point. I'm just a little confused how to use that to get the plane equation!

anonymous · Answer

well u have z=Ax+By+C for your plane am i right?

anonymous · Answer

the equations $\large  z_{x}=1 $ and $\large  z_{y}=-1 $ also must satisfy the plane

richyw · Answer

hmm. well usually I would do $z-z_0=f_x(x_0,y_0)(x-x_0)+f_y(x_0,y_0)(y-y_0)$ but with this implicit differentiation I am lost

richyw · Answer

so I think I see that that would be $z-z_0=(x-2)-(y-1)$. Is that correct so far? If so would the wohle thing just be$$z-1=x-y-1$$$$x-z-y=0$$

anonymous · Answer

thats right