# Thread: tangent plane problem (almost solved)

1. ## tangent plane problem (almost solved) [SOLVED]

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.

2. 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.
The original equation can be written as $\displaystyle 6x^2+3(y-z)^2-2y-2z=0$. Plug the information in red into that and it becomes $\displaystyle 3\bigl(\tfrac13\bigr)^2-2y-2\bigl(y-\tfrac13\bigr)=0$, which you can easily solve.

3. Thank you, I've got it.
But how did you recognize that the original equation could be rewritten like that?