Hyperboloids of One Sheet

Introduction

A hyperboloid of one sheet is a connected quadric surface that narrows to an elliptical waist and widens on either side. Its equation has a characteristic sign pattern: two positive squared terms and one negative squared term, with the equation equal to \(1\).

That sign pattern identifies the surface, but a useful sketch requires more. The negative term indicates the direction of the axis, the cross-section perpendicular to that axis at the center is the narrowest ellipse, and other carefully chosen sections reveal how the surface expands.

Standard Form and Axis

For a hyperboloid centered at the origin with axis parallel to the \(z\)-axis, the standard form is

The positive constants \(a\), \(b\), and \(c\) set the scale of the surface in the coordinate directions. The squared variable with the negative coefficient is \(z\), so the \(z\)-axis is the surface’s axis. If the negative term is instead \(x^2/c^2\) or \(y^2/c^2\), the axis is parallel to the \(x\)-axis or \(y\)-axis, respectively.

The phrase “one sheet” describes the connected surface represented by the equation. It does not mean that every coordinate-plane trace is connected: some traces are hyperbolas with two branches. The full surface joins around its narrowest elliptical cross-section.

The Narrowest Cross-Section

For the standard form with \(z\) as the axis, set \(z=0\). The equation becomes

This is an ellipse in the \(xy\)-plane with semiaxes \(a\) and \(b\). It is the waist, or narrowest cross-section, of the surface. In particular, when \(a\ne b\), the waist is not a circle.

To see how the perpendicular sections change, let \(z=k\), where \(k\) is any real number. Rearranging gives

The right-hand side is positive for every real \(k\). Dividing by it gives the ellipse in standard form:

Therefore the semiaxes of this section are \(a\sqrt{1+k^2/c^2}\) and \(b\sqrt{1+k^2/c^2}\). Since \(1+k^2/c^2\ge 1\), neither semiaxis is smaller than its waist value; both are smallest when \(k=0\). The sections at \(z=k\) and \(z=-k\) have the same size.

Why the Surface Is Connected

The cross-section calculation also explains why this is one connected surface rather than separate pieces. At every real value of \(z\), the equation describes an ellipse. No interval of \(z\)-values is excluded.

More explicitly, every point on the surface can be described using a height \(z\) and an angle \(\theta\):

Indeed, substituting these expressions into the left-hand side gives

Conversely, at any fixed \(z\), the cross-section is the ellipse just derived, and every point on that ellipse has such an angle \(\theta\). Thus the description covers the whole surface. From any point, hold \(\theta\) fixed and continuously move \(z\) to \(0\), changing \(x\) and \(y\) according to the displayed formulas. This gives a path along the surface to the waist. Points on the waist can be joined to one another by moving around its ellipse. Hence any two points on the surface can be connected by paths lying on the surface.

Hyperbolic Traces Through the Axis

Sections in planes containing the \(z\)-axis give a complementary view. Setting \(y=0\) produces the trace in the \(xz\)-plane:

This is a hyperbola opening in the positive and negative \(x\)-directions. Setting \(x=0\) produces the trace in the \(yz\)-plane:

a hyperbola opening in the positive and negative \(y\)-directions. Each trace passes through the waist at its corresponding endpoints: \((\pm a,0,0)\) in the \(xz\)-plane and \((0,\pm b,0)\) in the \(yz\)-plane.

These traces and the perpendicular ellipses serve different purposes. The ellipses show the surface’s changing width as one moves along its axis. The hyperbolas show its profile in two vertical planes. Together they provide a practical basis for sketching the three-dimensional surface.

Worked Examples

A Common Sign Mistake

Consider \(\frac{x^2}{4}+\frac{y^2}{9}-\frac{z^2}{16}=1\). A common error is to say that the axis is the \(x\)-axis because \(x^2\) appears first. The order of the terms has no geometric significance: the negative term is \(-z^2/16\), so the axis is the \(z\)-axis.

A second error is to conclude that the surface has no points at \(z=0\), perhaps because the negative term disappears. Substituting \(z=0\) correctly gives \(x^2/4+y^2/9=1\), an ellipse with real points such as \((2,0,0)\). Substitution confirms the correction: \(2^2/4+0-0=1\).