Originally Posted by

**superdude** question:Find the tangent plane that is horizontal to the surface Q:$\displaystyle 6x^2+3y^2+3z^2-6yz-2y-2z=0$. Hint this is the lowest point on the surface

work: I get the partial derivatives $\displaystyle f_x=\frac{-12x^2}{6z-6y-2}$ and $\displaystyle f_y=\frac{-6y-6z-2}{6z-6y-2}$. I then equate them with zero and from $\displaystyle f_x$ I get x=0 and from $\displaystyle f_y$ I get z=y-1/3

The formula of the tangent plane is $\displaystyle z=f(a,b)+f_x(a,b)(x-a)+f_y(a,b)(y-b)\Longrightarrow3y^2+3(y-\frac{1}{3})^2-6y^2-2y-2(y-\frac{1}{3})=0$ and solving for y I get $\displaystyle y=\pm\frac{sqrt{11}}{2}-\frac{3}{2}$ So now I'm stuck because I was thinking I could plug this into my original equation but I can't since I have two possibilities.