tangent plane problem (almost solved)

**superdude** · October 15th 2009, 07:51 PM

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

work: I get the partial derivatives $f_x=\frac{-12x^2}{6z-6y-2}$ and $f_y=\frac{-6y-6z-2}{6z-6y-2}$ . I then equate them with zero and from $f_x$ I get x=0 and from $f_y$ I get z=y-1/3
The formula of the tangent plane is $z=f(a,b)+f_x(a,b)(x-a)+f_y(a,b)(y-b)\Longrightarrow3y^2+3(y-\frac{1}{3})^2-6y^2-2y-2(y-\frac{1}{3})=0$ and solving for y I get $y=\pm\frac{\sqrt{11}}{2}-\frac{3}{2}$ 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.

**Opalg** · October 16th 2009, 12:26 AM

Originally Posted by superdude

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

work: I get the partial derivatives $f_x=\frac{-12x^2}{6z-6y-2}$ and $f_y=\frac{-6y-6z-2}{6z-6y-2}$ . I then equate them with zero and from $f_x$ I get x=0 and from $f_y$ I get z=y-1/3
The formula of the tangent plane is $z=f(a,b)+f_x(a,b)(x-a)+f_y(a,b)(y-b)\Longrightarrow3y^2+3(y-\frac{1}{3})^2-6y^2-2y-2(y-\frac{1}{3})=0$ and solving for y I get $y=\pm\frac{sqrt{11}}{2}-\frac{3}{2}$ 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 $6x^2+3(y-z)^2-2y-2z=0$ . Plug the information in red into that and it becomes $3\bigl(\tfrac13\bigr)^2-2y-2\bigl(y-\tfrac13\bigr)=0$ , which you can easily solve.

**superdude** · October 16th 2009, 10:06 PM

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

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