= Solution
Writing the coordinates as $(x,y,z)$, the level equation is $x^2+y^2-z^2=\alpha$. Therefore:
* \b[$\alpha=0$: a circular double cone], $x^2+y^2=z^2$. Its vertex at the origin is the sole singular point; every other point is a smooth surface point.
* \b[$\alpha=1$: a one-sheeted <hyperboloid>], with horizontal radius $\sqrt{1+z^2}$. It is smooth everywhere.
* \b[$\alpha=-1$: a two-sheeted <hyperboloid>], $z=\pm\sqrt{1+x^2+y^2}$. Both sheets are smooth everywhere, including their tips $(0,0,\pm1)$.
The cone vertex is an instance of the <double cone singularity>. The tips of the two-sheeted hyperboloid are regular because the <gradient> $(2x,2y,-2z)$ is nonzero there.
\Image[/past-exam-of-the-short-options-course-of-the-university-of-oxford/2017-quadratic-level-surfaces.png]
{title=Quadratic level surfaces: a singular double cone and two smooth hyperboloids}
{description=Original surface plots for levels zero, one and minus one. The red point marks the cone vertex, the only failure of the submanifold condition. The displayed hyperboloid tips are regular.}
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