# Thread: find horizontal tangent plane

1. ## find horizontal tangent plane

the surface $6x^2+3y^2+3z^2-6yz-2y-2z=0$ has one horizontal tangent plane. Find its equation and give the point where it is tangent.
Hint(?) there is no piont on the surface below this plane

I've taken the partial derivatives and got $\frac{dz}{dx}=\frac{-12x^2}{6z-6y-2}$ and $\frac{dz}{dy}=\frac{-6y-2}{6z-6y-2}$. I'm not sure what to do next since I can't equate them with 0 and solve for a variable.

I haven't learned about vectors yet

2. Originally Posted by superdude
the surface $6x^2+3y^2+3z^2-6yz-2y-2z=0$ has one horizontal tangent plane. Find its equation and give the point where it is tangent.
Hint(?) there is no piont on the surface below this plane

I've taken the partial derivatives and got $\frac{dz}{dx}=\frac{-12x^2}{6z-6y-2}$ and $\frac{dz}{dy}=\frac{-6y-2}{6z-6y-2}$. I'm not sure what to do next since I can't equate them with 0 and solve for a variable.

I haven't learned about vectors yet
Yes, you can. A horizontal plane must be of the form z= constant. Set the partial derivatives with respect to x and y equal to 0 and find the correct point of tangency.

3. I am unclear how to proceed. You want me to treat z as a constant but how do I know what to set it to?

$12x^2+6\frac{dz}{dx}-6y\frac{dz}{dx}=0$ and $6y+6\frac{dz}{dy}=6y\frac{dz}{dy}-2-2\frac{dz}{dy}=0$