Surface described by the equation $-3y^2 - 4xy + 2xz + 4yz - 2x - 2z + 1 = 0$

Solution 1:

Your idea is good but note that the equation

$$ -3y^2 - 4xy + 2xz + 4yz = 2x + 2z - 1 $$

does not define the intersection of the surface described by the quadratic form with the plane. Such an intersection would have been described by two equations and not a single equation:

$$ -3y^2 - 4xy + 2xz + 4yz = a, \,\,\, 2x + 2z - 1 = b. $$

What works better is to try and write your equation as $(\mathbf{x} - \mathbf{v})^T A (\mathbf{x} - \mathbf{v}) = c$. If $\mathbf{v} = \mathbf{0}$, you get the level set $q_A(\mathbf{x}) = \mathbf{x}^T A \mathbf{x} = c$ of the quadratic form associated to $A$ which you can analyze by your method. If $\mathbf{v} \neq 0$, the picture is the same with the "center" translated to $\mathbf{v}$ which is what I guess you call "the center of symmetry". Note that we have

$$ (\mathbf{x} - \mathbf{v})^T A (\mathbf{x} - \mathbf{v}) = \mathbf{x}^T A \mathbf{x} - \mathbf{v}^T A \mathbf{x} - \mathbf{x}^T A \mathbf{v} + \mathbf{v}^T \mathbf{A} \mathbf{v} = \mathbf{x}^T A \mathbf{x} - 2 \mathbf{v}^T A \mathbf{x} + \mathbf{v}^T \mathbf{A} \mathbf{v} $$

so the expression splits into a quadratic part in $\mathbf{x}$, a linear part and a constant part. In your case, quadratic part corresponds to:

$$ A = \begin{pmatrix} 0 & -2 & 1 \\ -2 & -3 & 2 \\ 1 & 2 & 0 \end{pmatrix}.$$

The linear part corresponds to

$$ -2 \mathbf{v}^T A \mathbf{x} = -2(v_1,v_2,v_3)A \begin{pmatrix} x \\ y \\ z \end{pmatrix} = -2 (v_1, v_2, v_3) \begin{pmatrix} -2y + z \\ -2x - 3y + 2z \\ x + 2y \end{pmatrix} = \\ -2v_1(-2y + z) -2v_2(-2x - 3y + 2z) -2v_3(x + 2y) = \\ x(4v_2 - 2v_3) + y(4v_1 + 6v_2 - 4v_3) -z(2v_1 +4v_2) \stackrel{\mbox{?}}{=} -2x - 2z $$

and so

$$ 4v_2 - 2v_3 = -2, \\ 4v_1 + 6v_2 - 4v_3 = 0,\\ 2v_1 + 4v_2 = 2 $$

which implies that

$$ \mathbf{v} = \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix}, \,\,\, \mathbf{v}^T A \mathbf{v} = 2. $$

Thus, we get

$$ (\mathbf{x} - \mathbf{v})^T A (\mathbf{x} - \mathbf{v}) = q_A(\mathbf{x} - \mathbf{v}) = -3y^2 -4xy + 2xz + 4yz -2x - 2z + 2. $$

To get your equation, we see that we need to choose $c = 1$ and then

$$ q_A(\mathbf{x} - \mathbf{v}) = -3y^2 -4xy + 2xz + 4yz -2x - 2z + 2 = 1 \iff \\ -3y^2 -4xy + 2xz + 4yz -2x - 2z + 1 = 0. $$

Since $A$ has eigenvalues $-5,1,1$, the surface is a one-sheeted circular hyperboloid (see here).