Introduction
A hyperboloid of two sheets consists of two separate pieces, or branches, positioned on opposite sides of a central gap. Its standard equation encodes where the branches point, where they begin, and what their cross-sections look like.
The key feature is the square with a positive coefficient on the left side of the equation: the variable in that term gives the direction of the axis. The two points where the branches begin are the vertices. Traces and cross-sections then provide the information needed for a sketch.
Standard Form and the Axis
With its center at the origin, a hyperboloid of two sheets has one of these standard forms:
where \(a,b,c>0\). In each equation, exactly one squared term is positive and the other two are negative. The positive term’s variable determines the axis: \(x\)-axis, \(y\)-axis, or \(z\)-axis, respectively. The denominator in that term determines the distance from the center to each vertex.
For example, in the first form the vertices are \((a,0,0)\) and \((-a,0,0)\). If the equation is centered at \((h,k,l)\), the corresponding form is
Its axis is parallel to the \(x\)-axis, and its vertices are \((h+a,k,l)\) and \((h-a,k,l)\). The same rule applies when the positive term contains \(y-k\) or \(z-l\) instead.
Why There Are Two Separate Branches
Consider the form with the \(x\)-axis as its axis. Rearranging its equation gives
Because squares are nonnegative, the right-hand side is at least \(1\). Therefore \(x^2/a^2\geq 1\), which means \(x\geq a\) or \(x\leq -a\). There are no points on the surface with \(-a<x<a\). The branches cannot join across this gap.
At \(x=a\) or \(x=-a\), the equation requires \(y=z=0\), giving the vertices. For \(x>a\) and for \(x<-a\), the surface continues outward. The same reasoning applies to the other standard forms, with \(y\) or \(z\) in place of \(x\).
Traces and Cross-Sections
A trace is the intersection of a surface with a plane. For the \(x\)-axis form centered at the origin, the coordinate-plane traces are found by setting one coordinate to zero:
- In the plane \(x=0\), the equation becomes \(-y^2/b^2-z^2/c^2=1\), which has no real solutions. This is the empty central trace across the gap.
- In the plane \(y=0\), the trace is \(x^2/a^2-z^2/c^2=1\), a hyperbola.
- In the plane \(z=0\), the trace is \(x^2/a^2-y^2/b^2=1\), also a hyperbola.
Cross-sections perpendicular to the axis show how the branches widen. Set \(x=t\), where \(t\) is a fixed number. The equation becomes
If \(|t|<a\), the right-hand side is negative and there is no cross-section. If \(t=\pm a\), the cross-section is a single vertex. If \(|t|>a\), the right-hand side is positive and the cross-section is an ellipse. Thus each branch begins at a vertex and widens as the distance from the center along the axis increases.
The axis changes which coordinate-plane trace is empty, but the method is the same: set the relevant coordinate equal to zero for a coordinate-plane trace, or equal to a constant for a cross-section. If the equation has a shifted center, use the plane through the center parallel to the relevant coordinate plane.
Worked Examples
Worked Example: Find the Axis, Vertices, and Traces
Analyze the surface
The positive term is \(x^2/9\), so the axis is the \(x\)-axis. Since \(a^2=9\), \(a=3\), and the vertices are \((3,0,0)\) and \((-3,0,0)\).
For the coordinate-plane traces, set each coordinate to zero in turn. If \(x=0\), then
The left side is at most zero, so this equation has no real solutions. If \(y=0\), the trace is \(x^2/9-z^2/16=1\). If \(z=0\), the trace is \(x^2/9-y^2/4=1\). These are hyperbolas.
As a cross-section check, set \(x=6\). Then \(x^2/9=36/9=4\), and the equation gives
This is an ellipse. The vertex check also works: at \((3,0,0)\), the left side is \(9/9-0-0=1\); at \((-3,0,0)\), it is again \(9/9-0-0=1\).
Worked Example: Analyze a Surface Centered Away from the Origin
Find the axis, vertices, and central-plane traces for
The positive term is \(y^2/25\), so the axis is parallel to the \(y\)-axis. The center is \((1,0,-2)\), and \(a=\sqrt{25}=5\). Moving \(5\) units in either direction along the \(y\)-axis from the center gives the vertices \((1,5,-2)\) and \((1,-5,-2)\).
The central plane perpendicular to the axis is \(y=0\). Substitution gives
There are no real points in this plane. The central plane \(x=1\), parallel to the \(yz\)-plane, gives \(y^2/25-(z+2)^2/9=1\). The central plane \(z=-2\), parallel to the \(xy\)-plane, gives \(y^2/25-(x-1)^2/4=1\). Both traces are hyperbolas.
To check the vertices, substitute \((1,5,-2)\): its left side is \(25/25-0-0=1\). Substituting \((1,-5,-2)\) gives the same value, since \((-5)^2/25=1\). For an additional cross-section, set \(y=10\). Then \(y^2/25=100/25=4\), so
This is an ellipse in the plane \(y=10\), confirming that a cross-section beyond either vertex is a closed curve.
Worked Example: Sketch Using Traces and a Cross-Section
Analyze the surface
The positive term is \(z^2/4\), so the axis is the \(z\)-axis. Here \(a=2\), and the vertices are \((0,0,2)\) and \((0,0,-2)\). The surface can only occur where \(z\geq 2\) or \(z\leq -2\), so the sketch must have two branches separated by a gap around the origin.
Setting \(z=0\) gives \(-x^2/9-y^2/16=1\), which has no solutions. Setting \(y=0\) gives the \(xz\)-plane trace \(z^2/4-x^2/9=1\). Setting \(x=0\) gives the \(yz\)-plane trace \(z^2/4-y^2/16=1\). Both nonempty traces are hyperbolas opening in the positive and negative \(z\)-directions.
Now set \(z=4\). Since \(z^2/4=16/4=4\), the cross-section equation is
This is an ellipse. For a direct check, the point \((3\sqrt{3},0,4)\) satisfies the original equation: \(z^2/4=4\), \(x^2/9=(9\cdot3)/9=3\), and \(y^2/16=0\), so the left side is \(4-3-0=1\). The traces and this ellipse show how to draw two branches starting at \(z=\pm2\) and widening as they extend away from the center.
A Reliable Sketching Procedure
For a surface already written in standard form, use the following sequence. The axis term identifies the direction first; the other features provide checks on the sketch.
The variable in the single positive square gives the direction of the axis.
The center comes from the shifts in the squared terms; the positive term’s denominator gives the square of the distance to each vertex.
Set the axis coordinate equal to its center value to see the gap, then set each other coordinate equal to its center value to obtain the two hyperbolic traces.
Fix the axis coordinate farther from the center. The resulting ellipse shows how a branch widens.
Start at the vertices and extend away from the central gap, consistent with the traces and cross-sections.
Common Mistake: Using a Negative Term to Choose the Axis
For a hyperboloid of two sheets, the axis is not selected by choosing one of the negative terms. For example, in
a mistaken reading might point to \(y\) or \(z\) because those terms have negative signs. That would give the wrong direction. The positive term is \(x^2/9\), so the axis is the \(x\)-axis and the vertices are \((\pm3,0,0)\). Setting \(x=0\) confirms the gap: the resulting equation is \(-y^2/4-z^2/16=1\), which is impossible over the real numbers.
This sign rule is specific to the two-sheet form described here: one positive square and two negative squares equal to \(1\). Check the full standard form before identifying the axis.
Why the Traces Matter
A three-dimensional equation is difficult to picture from its symbols alone. Traces turn it into familiar two-dimensional curves. In the central plane perpendicular to the axis, the trace is empty; in the other two central coordinate planes, the traces are hyperbolas. Cross-sections farther along the axis are ellipses. Together, these facts locate the gap, show the direction of the branches, and indicate how they widen.
These checks also help catch a mismatched sketch. A drawing that passes through the empty central trace, joins the two branches, or narrows toward the vertices instead of widening does not match the standard form.
Check Your Understanding
Use standard form and traces to answer the following questions.
- For \(y^2/16-x^2/9-z^2/4=1\), what is the axis, and where are the vertices?
- For \(x^2/25-y^2/4-z^2/9=1\), what equation describes the trace in the plane \(z=0\)?
- For \(z^2/9-x^2/4-y^2/16=1\), does the plane \(z=0\) intersect the surface? Explain using the equation.
- In \(x^2/4-y^2/9-z^2/16=1\), what kind of cross-section results from setting \(x=4\)?
- Why can a hyperboloid of two sheets not connect one branch to the other?