What surface is $x^2-4xy-2xz+z^2=1$?
I just convert the equation to $(x-2y)^2+(x-z)^2=1+x^2+4y^2$ and $2(x-y)^2+(x-z)^2 = 1+2x^2+2y^2$.
What is standard method to solve this kind of problem?
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Get it into $$A x^2 + B y ^2 + C z^2 + 2 D x y + 2 E y z + 2 F z x + 2 G x + 2 H y + 2 I z + J = 0$$ form – John Alexiou Jul 18 '17 at 23:00
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and then have a look at this article and to this other post – G Cab Jul 18 '17 at 23:34
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A standard method is to write your equation as $x^{T}Ax=c$, where $x$ is a vector, $A$ is a symmetric matrix, and $c$ is a constant.
If we let $A=\begin{bmatrix} 1 & -2 & -1 \\ -2 & 0 & 0 \\ -1 & 0 & 1 \end{bmatrix}$, $x=\begin{bmatrix} x \\ y \\ z \end{bmatrix}$, and $c=1$ that does the trick. Then look at the eigenvalues of $A$.
The idea is to change from the $xyz$ coordinate system with orthogonal basis vectors to another coordinate system with orthogonal basis vectors to understand our object better. To do this perform the change of variables $x=Py$ where $P$ is some orthogonal matrix. Also, diagonalizing $A$ into $\color{red}{PDP^{T}}$ will prove to be useful.
$$(Py)^{T}\color{red}{PDP^{T}}(Py)=c$$
$$y^{T}P^{T}\color{red}{PDP^{T}}Py=c$$
Orthogonal matrices have the property that $P^{T}P=I$ so we get,
$$y^TDy=c$$
That equates to,
$$y_1^2\lambda_1+y_2^2\lambda_2+....+y_n^2 \lambda_n=c$$
In the $y_1y_2...y_n$ coordinate system. Where $\lambda_i$ are the eigenvalues of $A$.
Here the eigenvalues are messy, satisfying $-\lambda^3+2\lambda^2+4\lambda-4=0$. But it's not that hard to see two are positive and and another is negative. That tells me we have a Hyperboloid of One Sheet.
Ahmed S. Attaalla
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