Q. \(x^2 + y^2 – z^2 = 1\) hyperboloid of one sheet.

Problem

Analyze and describe the surface given by

\[ x^{2} + y^{2} – z^{2} = 1 \]

1. Identify the type of surface

   Compare the equation with the standard quadratic surface forms. The left-hand side has two positive squared terms and one negative squared term, and the right-hand side is a positive constant. This is the standard equation of a hyperboloid of one sheet. In standard form it can be read as

   \[ \frac{x^{2}}{1} + \frac{y^{2}}{1} – \frac{z^{2}}{1} = 1. \]

   Because exactly one squared term has the opposite sign from the other two and the constant on the right-hand side is positive, the surface is a hyperboloid of one sheet.

2. Symmetries

   From the equation it is clear that the surface is symmetric under:

   • Rotation about the \(z\)-axis, because \(x\) and \(y\) appear only as \(x^{2} + y^{2}\).
   • Reflection through the coordinate planes, because only even powers appear.
   • Central symmetry through the origin: if \((x,y,z)\) is on the surface then \((-x,-y,-z)\) is also on the surface.

3. Horizontal (constant z) cross sections — circles

   Take a horizontal plane \(z = z_{0}\). Substitute into the equation:

   \[ x^{2} + y^{2} = 1 + z_{0}^{2}. \]

   This is a circle centered at the origin in the \(xy\)-plane with radius

   \[ r(z_{0}) = \sqrt{1 + z_{0}^{2}}. \]

   Because \(1 + z_{0}^{2} > 0\) for all real \(z_{0}\), every horizontal slice is a nondegenerate circle. At \(z = 0\) the radius is \(1\); as \(|z|\) increases the radius grows without bound. This continuous family of circles stacked along \(z\) is the reason the surface is a single connected sheet.

4. Vertical cross sections (x = constant or y = constant) — hyperbolas

   Fix \(x = x_{0}\) and substitute into the equation:

   \[ y^{2} – z^{2} = 1 – x_{0}^{2}. \]

   Depending on the value of \(x_{0}\):

   • If \(|x_{0}| < 1\) then the right-hand side is positive and the slice is a standard rectangular hyperbola with transverse axis along the \(y\)-direction.
   • If \(|x_{0}| = 1\) then the right-hand side is zero and the slice degenerates into the pair of lines \(y = \pm z\).
   • If \(|x_{0}| > 1\) then the right-hand side is negative; rewriting as \(z^{2} – y^{2} = x_{0}^{2} – 1\) shows again a hyperbola (now with transverse axis along \(z\)).

   Similarly, fixing \(y = y_{0}\) yields slices \(x^{2} – z^{2} = 1 – y_{0}^{2}\), also hyperbolas (or degenerate lines) for various \(y_{0}\). These hyperbolic vertical traces, together with circular horizontal traces, are characteristic of the one-sheet hyperboloid.

5. Asymptotic cone

   For large distance from the origin the constant \(1\) becomes negligible, so the surface approaches the cone given by setting the right-hand side to zero:

   \[ x^{2} + y^{2} – z^{2} = 0. \]

   Equivalently, \(z = \pm\sqrt{x^{2} + y^{2}}\). This double cone is the asymptotic cone of the hyperboloid: far from the origin the hyperboloid gets closer and closer to this cone.

6. Parametrizations

   Two convenient parametrizations are given below.

   1) Hyperbolic-trigonometric (global, smooth): Let \(u \in \mathbb{R}\) and \(v \in [0,2\pi)\). Then

   \[ x(u,v) = \cosh u \cos v, \quad y(u,v) = \cosh u \sin v, \quad z(u,v) = \sinh u. \]

   Verify by substitution:

   \[ x^{2} + y^{2} – z^{2} = \cosh^{2}u(\cos^{2}v+\sin^{2}v) – \sinh^{2}u = \cosh^{2}u – \sinh^{2}u = 1. \]

   This parametrization covers the entire surface once as \(u\) runs over all real numbers and \(v\) runs from \(0\) to \(2\pi\).

   2) Cylindrical-style (useful to visualize horizontal circles): Let \(\theta \in [0,2\pi)\) and \(z \in \mathbb{R}\). Then

   \[ x(\theta,z) = \sqrt{1+z^{2}}\cos\theta, \quad y(\theta,z) = \sqrt{1+z^{2}}\sin\theta, \quad z(\theta,z) = z. \]

   Substituting shows

   \[ x^{2} + y^{2} – z^{2} = (1+z^{2}) – z^{2} = 1. \]

   This parametrization emphasizes the circles at fixed \(z\).

7. Sketching guidance

   Key steps to sketch the surface:

   1. Draw the circle at \(z = 0\) of radius \(1\) in the \(xy\)-plane.
   2. For several values \(z = \pm 1, \pm 2\), draw circles of radii \(\sqrt{1+z^{2}}\) (for \(z = \pm 1\) radius \(\sqrt{2}\), for \(z = \pm 2\) radius \(\sqrt{5}\), etc.).
   3. Connect these circles smoothly to show a single continuous “waist” at \(z = 0\) expanding outward as \(|z|\) increases.
   4. Optionally draw vertical hyperbolic traces in planes \(x = 0\) and \(y = 0\) to show the hyperbolic cross sections.
   5. Draw the asymptotic cone \(z = \pm\sqrt{x^{2}+y^{2}}\) as a guide; the hyperboloid approaches this cone at large radius.

8. Summary

   The surface \[ x^{2} + y^{2} – z^{2} = 1 \] is a hyperboloid of one sheet. It is rotationally symmetric about the \(z\)-axis, has circular horizontal cross sections of radius \(\sqrt{1+z^{2}}\), hyperbolic vertical cross sections, and approaches the cone \[ x^{2} + y^{2} – z^{2} = 0 \] at infinity. A convenient global parametrization is

   \[ x = \cosh u \cos v, \quad y = \cosh u \sin v, \quad z = \sinh u, \quad u\in\mathbb{R},\ v\in[0,2\pi). \]