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Coordinates and Quadric Surfaces · Tutorial 12 of 999

Hyperbolic Paraboloids

Learn to read the orientation of a hyperbolic paraboloid from its equation and assemble its shape from horizontal and vertical traces.

Beginner 10 min read

What You'll Learn

  • Identify the standard equations of hyperbolic paraboloids.
  • Determine which coordinate directions rise and which fall.
  • Derive horizontal hyperbola sections for positive, negative, and zero heights.
  • Use vertical parabolas to confirm the saddle’s orientation.
  • Sketch a surface by combining its key cross-sections.
  • Catch sign errors when rewriting horizontal sections.

Introduction

A hyperbolic paraboloid is a familiar saddle shape: it curves upward in one coordinate direction and downward in another. The equation may look like a difference of squares, but the signs alone are only the start. To sketch the surface reliably, connect those signs to the vertical parabolas and horizontal hyperbolas cut out by coordinate planes and horizontal planes.

In this tutorial, the \(z\)-axis is treated as vertical. A horizontal cross-section lies in a plane \(z=k\); vertical cross-sections lie in planes such as \(x=0\) or \(y=0\). Together, these traces show which way the saddle opens and how its shape changes with height.

Standard Forms and Orientation

The two basic forms are obtained by subtracting one squared coordinate term from another. The positive coefficients determine the scale in each direction, while the minus sign determines which direction curves downward.

Definition: A hyperbolic paraboloid in standard position has one of the forms
$$ z=\frac{x^2}{a^2}-\frac{y^2}{b^2} \qquad\text{or}\qquad z=\frac{y^2}{b^2}-\frac{x^2}{a^2}, \quad a>0,\ b>0. $$
The origin is the saddle point. In the first form the surface curves upward along the \(x\)-direction and downward along the \(y\)-direction; in the second form those roles are reversed.

For the first equation, setting \(y=0\) leaves \(z=x^2/a^2\), a parabola opening upward. Setting \(x=0\) leaves \(z=-y^2/b^2\), a parabola opening downward. The second equation reverses these vertical traces. These opposing parabolas explain the saddle shape: moving away from the origin in one direction raises the surface, while moving away in the perpendicular direction lowers it.

The parameters \(a\) and \(b\) are positive, so their squares do not change the signs. They control the widths of the cross-sections. The equation has no nonzero constant term, so the surface passes through the origin; substituting \(x=y=0\) gives \(z=0\).

Horizontal Cross-Sections

To find a horizontal trace, hold the height fixed by setting \(z=k\). For the first standard form, this gives an equation in \(x\) and \(y\). The sign of \(k\) determines which coordinate direction is the transverse direction of the hyperbola: when \(k\) is positive, the hyperbola opens along the \(x\)-axis; when \(k\) is negative, it opens along the \(y\)-axis.

Result: For \(z=x^2/a^2-y^2/b^2\), the horizontal trace at height \(k\) is
$$ \begin{aligned} k>0:\quad& \frac{x^2}{a^2k}-\frac{y^2}{b^2k}=1,\\[4pt] k<0:\quad& \frac{y^2}{b^2|k|}-\frac{x^2}{a^2|k|}=1. \end{aligned} $$
At \(k=0\), the trace is a pair of intersecting lines, not a hyperbola.

Here is the sign-sensitive step for a negative height. Write \(k=-|k|\), then rearrange the equation before dividing:

$$ \frac{x^2}{a^2}-\frac{y^2}{b^2}=-|k| \quad\Longleftrightarrow\quad \frac{y^2}{b^2}-\frac{x^2}{a^2}=|k| \quad\Longleftrightarrow\quad \frac{y^2}{b^2|k|}-\frac{x^2}{a^2|k|}=1. $$

The rearrangement puts the positive term first; dividing by the positive number \(|k|\) then gives the standard hyperbola equation. When \(k=0\), instead factor the difference of squares:

$$ \frac{x^2}{a^2}-\frac{y^2}{b^2}=0 \quad\Longleftrightarrow\quad \left(\frac{x}{a}-\frac{y}{b}\right) \left(\frac{x}{a}+\frac{y}{b}\right)=0. $$

Thus the zero-height trace consists of the two lines \(y=(b/a)x\) and \(y=-(b/a)x\). They meet at the saddle point.

Why the Horizontal Traces Have These Orientations

For a positive height, the horizontal-section equation is \(x^2/a^2-y^2/b^2=k\), where \(k>0\). Dividing by \(k\) yields a positive \(x^2\) term equal to \(1\) plus a nonnegative \(y^2\) term. Setting \(y=0\) gives \(x^2=a^2k\), so the vertices are \((x,y)=(\pm a\sqrt{k},0)\). This identifies the \(x\)-axis as the direction in which the hyperbola opens.

For a negative height, the same equation can be rearranged as \(y^2/b^2-x^2/a^2=|k|\). Dividing by \(|k|\) gives a positive \(y^2\) term equal to \(1\) plus a nonnegative \(x^2\) term. Setting \(x=0\) gives \(y^2=b^2|k|\), so the vertices are \((0,\pm b\sqrt{|k|})\). The hyperbola therefore opens along the \(y\)-axis. This sign change is why the horizontal sections switch direction as the height passes through zero.

For \(z=y^2/b^2-x^2/a^2\), the same reasoning applies with \(x\) and \(y\) exchanged: positive heights give hyperbolas opening along \(y\), and negative heights give hyperbolas opening along \(x\).

Worked Examples

Worked Example: Sketching a Saddle That Rises Along the \(x\)-Direction

Consider \(z=x^2/4-y^2/9\). First, set one horizontal coordinate to zero at a time. In the plane \(y=0\), the trace is \(z=x^2/4\), which opens upward. In the plane \(x=0\), the trace is \(z=-y^2/9\), which opens downward. The surface therefore rises along \(x\) and falls along \(y\).

At height \(z=4\), substitute \(k=4\):

$$ \frac{x^2}{4}-\frac{y^2}{9}=4 \quad\Longleftrightarrow\quad \frac{x^2}{16}-\frac{y^2}{36}=1. $$

This hyperbola opens along the \(x\)-axis. Its vertices are \((4,0)\) and \((-4,0)\) in the horizontal plane. Check the first vertex in the surface equation: \(z=4^2/4-0^2/9=16/4=4\). At the other vertex, \(z=(-4)^2/4-0^2/9=16/4=4\).

At height \(z=-9\), substitute \(k=-9\) and rearrange:

$$ \frac{x^2}{4}-\frac{y^2}{9}=-9 \quad\Longleftrightarrow\quad \frac{y^2}{9}-\frac{x^2}{4}=9 \quad\Longleftrightarrow\quad \frac{y^2}{81}-\frac{x^2}{36}=1. $$

This hyperbola opens along the \(y\)-axis, with vertices \((0,9)\) and \((0,-9)\). Checking them in the original equation gives \(z=0^2/4-9^2/9=-81/9=-9\) and \(z=0^2/4-(-9)^2/9=-81/9=-9\). At height zero, the trace is \(x^2/4-y^2/9=0\), or \(3x=2y\) and \(3x=-2y\). These traces together give the surface’s saddle orientation and changing horizontal direction.

Worked Example: Sketching the Reversed Orientation

Consider \(z=y^2/16-x^2/9\). The vertical trace in \(x=0\) is \(z=y^2/16\), which opens upward. The trace in \(y=0\) is \(z=-x^2/9\), which opens downward. This time the surface rises along \(y\) and falls along \(x\).

At height \(z=4\), the horizontal trace is

$$ \frac{y^2}{16}-\frac{x^2}{9}=4 \quad\Longleftrightarrow\quad \frac{y^2}{64}-\frac{x^2}{36}=1. $$

Its vertices are \((x,y)=(0,8)\) and \((0,-8)\), so it opens along \(y\). Substitution confirms the height: at either vertex, \(z=8^2/16-0^2/9=64/16=4\).

At height \(z=-1\), rearrange rather than leaving the negative right-hand side in place:

$$ \frac{y^2}{16}-\frac{x^2}{9}=-1 \quad\Longleftrightarrow\quad \frac{x^2}{9}-\frac{y^2}{16}=1. $$

This hyperbola opens along \(x\), with vertices \((3,0)\) and \((-3,0)\). Each gives \(z=0^2/16-3^2/9=-9/9=-1\), as required. The positive- and negative-height traces open in opposite directions, consistent with the vertical parabolas.

Worked Example: Reading the Shape from Unequal Coefficients

Sketch \(z=3x^2-2y^2\). The coefficient of \(x^2\) is positive and the coefficient of \(y^2\) is negative. Setting \(y=0\) gives \(z=3x^2\), an upward-opening parabola; setting \(x=0\) gives \(z=-2y^2\), a downward-opening parabola.

At height \(z=6\), solve for the horizontal trace and divide by \(6\):

$$ 3x^2-2y^2=6 \quad\Longleftrightarrow\quad \frac{x^2}{2}-\frac{y^2}{3}=1. $$

The hyperbola opens along \(x\), with vertices \((\sqrt{2},0)\) and \((-\sqrt{2},0)\). Substituting either \(x\)-value and \(y=0\) into the surface equation gives \(z=3(\sqrt{2})^2=3\cdot2=6\), or \(z=3(-\sqrt{2})^2=6\).

At height \(z=-8\), rearrange to put the positive term first:

$$ 3x^2-2y^2=-8 \quad\Longleftrightarrow\quad 2y^2-3x^2=8 \quad\Longleftrightarrow\quad \frac{y^2}{4}-\frac{x^2}{8/3}=1. $$

The hyperbola opens along \(y\), with vertices \((0,2)\) and \((0,-2)\). Substitution gives \(z=3(0)^2-2(2)^2=-8\) and \(z=3(0)^2-2(-2)^2=-8\). These computed traces and the two vertical parabolas are enough to place the surface’s principal directions and sketch its saddle shape.

A Reliable Sketching Process

A sketch does not need to show every point on the surface. It should make the saddle point, the vertical directions of curvature, and the change in horizontal hyperbola orientation clear. For an equation in standard position, use the following sequence.

1
Read the signs.
Identify which squared term is positive and which is negative. The positive direction curves upward; the negative direction curves downward.
2
Draw the vertical traces.
Set \(x=0\) and \(y=0\) in turn. These parabolas verify the direction of curvature along both coordinate axes.
3
Choose positive and negative heights.
Set \(z=k\) for one \(k>0\) and one \(k<0\). Put the positive squared term first and divide by the positive magnitude of the height.
4
Mark vertices and the zero trace.
Use the hyperbola vertices to show the opening directions. At \(z=0\), show the intersecting lines through the saddle point.

Common Sign Error

A frequent mistake occurs when the height is negative: the equation is rearranged without changing the signs correctly. For example, on the surface \(z=x^2-y^2\), the section \(z=-1\) satisfies

$$ x^2-y^2=-1 \quad\Longleftrightarrow\quad y^2-x^2=1, $$

not \(y^2-x^2=-1\). The corrected equation has vertices \((0,1)\) and \((0,-1)\), as direct substitution confirms: \(0^2-1^2=-1\). The incorrect equation would place its vertices on the \(x\)-axis and predict the wrong opening direction. A useful check is to substitute a simple point from the claimed trace into the original equation before using it in a sketch.

Why the Saddle Description Matters

The vertical traces and horizontal traces give independent checks on the same surface. The vertical parabolas show which coordinate directions rise and fall. The horizontal hyperbolas show that the direction of their vertices changes as the height changes sign. If a proposed sketch has horizontal hyperbolas opening in the same direction at positive and negative heights, or has both vertical traces opening upward, it cannot represent either standard hyperbolic paraboloid.

The distinction also prevents confusing a saddle with a surface that curves in the same vertical direction along both coordinate axes. The defining feature here is not simply that some sections are hyperbolas: it is the coordinated pattern of oppositely opening vertical parabolas and horizontal hyperbolas whose orientation switches across height zero.

Check Your Understanding

Use the equations and cross-section method in this tutorial to answer the following.

  1. For \(z=x^2/9-y^2/4\), which vertical trace opens upward, and which opens downward?
  2. For that same surface, in which coordinate direction does the horizontal hyperbola at a positive height open?
  3. What is the horizontal trace at height zero for \(z=y^2/25-x^2/16\)?
  4. At height \(z=-4\), rewrite \(x^2/4-y^2/9=-4\) as a standard hyperbola equation. Which axis does it open along?
  5. What two features of a sketch would you check to make sure the saddle orientation is consistent with its equation?

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