Name the Surface

The equation is invariant under $x \mapsto -x$ , $y \mapsto -y$ , $z \mapsto -z$ . This point symmetry excludes the paraboloid.

The equation is invariant under rotation of the variables $x\mapsto y\mapsto z\mapsto x$ . Therefore the line through the origin in direction $(1,1,1)$ is an axis of rotational symmetry; the rotation of variables is a rotation of $120^\circ$ about this axis.

As we travel from the origin along this line, i.e. $(x,y,z) \to (+\infty,+\infty,+\infty)$ , we see that $xy + xz + yz$ increases from zero to infinity. It will reach the value 1 at some point on this line; therefore the surface is not empty or degenerate.

Consider the plane through the origin perpendicular to the line, i.e. the plane $x + y + z = 0$ . On this entire plane we have $xy + xz + yz = \tfrac12(x+y+z)^2 - (x^2+y^2+z^2) = -(x^2 + y^2 + z^2) \leq 0.$ It follows that the surface does not intersect this entire plane through the origin. This shows that the surface consists of two branches, or sheets.

This is enough to conclude it is a $\boxed{\text{hyperboloid of two sheets}}$ .

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Alternative

The equation may be written in the form $\left(\begin{array}{ccc} x & y & z \end{array}\right)\:M\:\left(\begin{array}{c} x \\ y \\ z\end{array}\right) = 1,$ where $M$ is a symmetric matrix, namely $M = \left(\begin{array}{ccc} 0 & \tfrac12 & \tfrac12 \\ \tfrac12 & 0 & \tfrac12 \\ \tfrac12 & \tfrac12 & 0\end{array}\right).$ To find the eigenvalues of $M$ , we solve $0 = \text{det}\ (M - \lambda I) = \lambda^3 - \tfrac34\lambda - \tfrac14$ , giving $\lambda = 1$ (once) and $\lambda = -\tfrac12$ (twice). At this point we might already draw the conclusion that the surface is a hyperboloid of two sheets, based on the eigenvalue "signature":

• $(+,+,+)$ corresponds to $a^2 + b^2 + c^2 = 1$ , an ellipsoid;

• $(+,+,-)$ corresponds to $a^2 + b^2 = 1 + c^2$ , a one-sheeted hyperboloid;

• $(-,-,+)$ corresponds to $a^2 + b^2 = -1 + c^2$ , a two-sheeted hyperboloid;

• $(-,-,-)$ corresponds to $a^2 + b^2 + c^2 = -1$ , an empty surface;

• one or more zero eigenvalues correspond to paraboloids and degenerate surfaces.

To continue in more detail, a set of orthogonal eigenvectors include $\left(\begin{array}{c} 1 \\ 1 \\ 1 \end{array}\right),\ \ \ \ \left(\begin{array}{c} 2 \\ -1 \\ -1\end{array}\right),\ \ \ \ \left(\begin{array}{c} 0 \\ 1 \\ -1\end{array}\right).$ This suggests the change of coordinates $\begin{cases} a = \tfrac13(x+y+z) & x = a + 2b \\ b = \tfrac16(2x - y - z) & y = a - b + c \\ c = \tfrac12(y - z) & z = a - b - c\end{cases}$ The equation becomes $1 = xy + yz + xz = (a + 2b)(a - b + c) + (a - b + c)(a - b - c) + (a + 2b)(a - b - c) = 2(a + 2b)(a-b) + (a-b)^2 - c^2 = 3a^2 - 3b^2 - c^2.$ We may write this as $3b^2 + c^2 = 3a^2 - 1;$ for every value of $|a| > 1/\sqrt{3}$ the left-hand side describes an ellipse; for every value of $|a| < 1/\sqrt{3}$ this equation has no solutions. Thus, when we slice the surface perpendicular to $\vec a$ , we find a stack of growing ellipses on either side, with a gap around the origin: a hyperboloid of two sheets.