1. ## Equation Manipulation

I'm stuck dealing with a 3 variable equation and I am trying to determine what kind of 3D surface is produced. the equation is:

x + z^2 + y^2 + 6zy = 0

I have done traces and attempted to draw the surface but I need to manipulate the equation to look like one of the standard forms we know. From graphing software I believe the surface created is a hyperbolic paraboloid which means the above equation should look something like...

z/c = x^2/a^2 - y^2/b^2 which is the standard form of the hyperbolic.

Ive tried completing the square to get rid of the xy term but still end up with an equation that is not close to the standard form. Any suggestions would be greatly appreciated!

2. Here is the way I know how to do it..

Consider the general form of the quadric: $\displaystyle a_{11}x^2+a_{22}y^2+a_{33}z^2+2a_{12}xy+2a_{13}xz+ 2a_{23}yz+2a_{14}x+2a_{24}y+2a_{34}z+a_{44}=0$.
Denote $\displaystyle a=\begin{pmatrix}a_{11}&a_{12}&a_{13}\\a_{12}&a_{2 2}&a_{23}\\a_{13}&a_{23}&a_{33}\end{pmatrix}$ and $\displaystyle A=\begin{pmatrix}a_{11}&a_{12}&a_{13}&a_{14}\\a_{1 2}&a_{22}&a_{23}&a_{24}\\a_{13}&a_{23}&a_{33}&a_{3 4}\\a_{14}&a_{24}&a_{34}&a_{44}\end{pmatrix}$
You have $\displaystyle a=\begin{pmatrix}0&0&0\\0&1&3\\0&3&1\end{pmatrix}$ with $\displaystyle \det a=0$ (so the quadric is without unique centre) and $\displaystyle A=\begin{pmatrix}0&0&0&{1/2}\\0&1&3&0\\0&3&1&0\\{1/2}&0&0&0\end{pmatrix}$ with $\displaystyle \det A=2\neq0$ (so the quadric in non-degenerated).
Hence, it can only be an elyptic paraboloid or a hyperbolic paraboloid.

The roots of the characteristic polynomial $\displaystyle p(\lambda)=\det(a-\lambda I_3)$ are $\displaystyle \lambda_1=0,\lambda_2=-2,\lambda_3=4$. The sets of the respective eigenvectors are $\displaystyle S(\lambda_1)=(1,0,0)\mathbb{R},\,S(\lambda_2)=(0,1 ,-1)\mathbb{R},\,S(\lambda_3)=(0,1,1)\mathbb{R}$.
Denote $\displaystyle v_1=(1,0,0),\,v_2=(0,1,-1),\,v_3=(0,1,1)$. They are already orthogonalized, so we only normalize them: $\displaystyle f_1:=v_1,f_2:=\frac{1}{\sqrt{2}}v_2,f_3:=\frac{1}{ \sqrt{2}}v_3$ $\displaystyle \Longrightarrow \{f_1,f_2,f_3\}$ is a orthonormal basis in $\displaystyle \mathbb{R}^3\Rightarrow$ the next transformation is an isometry:
$\displaystyle \begin{cases}x=1\cdot X+0\cdot Y+0\cdot Z\\y=0\cdot X+\frac{1}{\sqrt{2}}\cdot Y+\frac{1}{\sqrt{2}}\cdot Z\\x=0\cdot X-\frac{1}{\sqrt{2}}\cdot Y+\frac{1}{\sqrt{2}}\cdot Z\end{cases}$
By that, the quadric becomes: $\displaystyle X-2Y^2+4Z^2=0$, which is indeed a hyperbolic paraboloid, like you had noticed already

I guess there could be a simpler way though, but I don't see it now..