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Coordinates and Quadric Surfaces · Tutorial 24 of 999

Comparing Quadrics Through Horizontal Cross-Sections

Learn to read how horizontal ellipses change with height and use that pattern to distinguish common quadric surfaces.

Beginner 9 min read

What You'll Learn

  • Define a horizontal cross-section by intersecting a surface with a plane of constant height.
  • Find the ellipse in a horizontal slice by substituting the height into a quadric equation.
  • Track how the semiaxes of ellipsoids and cones vary with height.
  • Distinguish one-sheet and two-sheet hyperboloids using slice size and allowed heights.
  • Identify a quadric from a sequence of horizontal cross-sections.
  • Avoid confusing expanding sections with sections that exist only beyond a threshold height.

Reading a Surface One Height at a Time

A three-dimensional surface can be difficult to recognize from its equation alone. One useful strategy is to examine what remains when the surface is cut by horizontal planes. Each plane has a fixed height, so its intersection with the surface is a two-dimensional curve. As the height changes, the curves reveal how the surface is organized.

For ellipsoids, cones, and hyperboloids whose axes are parallel to the \(z\)-axis, these horizontal intersections are typically ellipses, with a point possible at a tip or boundary. The key is not just to notice that the slices are elliptical. Track whether the ellipses shrink, grow, or appear only above or below a certain height.

Definition: A horizontal cross-section of a surface at height \(k\) is its intersection with the plane \(z=k\). To find it algebraically, substitute \(k\) for \(z\) in the surface equation and interpret the resulting equation in \(x\) and \(y\).

How to Read the Ellipse in a Slice

An equation of the form

$$ \frac{x^2}{A^2}+\frac{y^2}{B^2}=1 $$

describes an ellipse in the horizontal plane, provided \(A^2>0\) and \(B^2>0\). Its semiaxis lengths are \(A\) in the \(x\)-direction and \(B\) in the \(y\)-direction, where \(A\) and \(B\) are positive. If the right-hand side is a positive number \(q\), then

$$ \frac{x^2}{a^2}+\frac{y^2}{b^2}=q \quad\Longleftrightarrow\quad \frac{x^2}{a^2q}+\frac{y^2}{b^2q}=1. $$

Thus the semiaxis lengths are \(a\sqrt{q}\) and \(b\sqrt{q}\). The same scale factor \(\sqrt{q}\) changes both semiaxes. If \(q=0\), the equation reduces to \(x=y=0\), a single point. If \(q<0\), there are no real points because a sum of nonnegative terms cannot be negative.

This gives a consistent way to compare surfaces: at each height, find the factor on the right-hand side of the normalized ellipse equation. Then determine whether that factor is positive, zero, or negative and how it changes with height.

Ellipsoids: Largest at the Middle

Consider an ellipsoid centered at the origin with its principal axis along \(z\):

$$ \frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1, \qquad a,b,c>0. $$

At height \(z=k\), substitution gives

$$ \frac{x^2}{a^2}+\frac{y^2}{b^2} =1-\frac{k^2}{c^2}. $$

The right-hand side is positive only when \(-c<k<c\). At \(k=0\), the horizontal ellipse has semiaxes \(a\) and \(b\), so it is the largest horizontal slice. As \(|k|\) increases toward \(c\), the right-hand side decreases toward zero and both semiaxes shrink. At \(k=c\) or \(k=-c\), the slice is a point; for \(|k|>c\), there is no intersection.

Worked Example: Tracking Ellipsoid Slices

Find the horizontal cross-sections at \(z=0\), \(z=2\), and \(z=3\) for

$$ \frac{x^2}{9}+\frac{y^2}{4}+\frac{z^2}{9}=1. $$

At \(z=0\), substitute \(0\) for \(z\):

$$ \frac{x^2}{9}+\frac{y^2}{4}=1. $$

This is an ellipse with semiaxes \(3\) and \(2\). Check: setting \(x=3,y=0,z=0\) gives \(9/9+0+0=1\); setting \(x=0,y=2,z=0\) gives \(0+4/4+0=1\).

At \(z=2\), substitute and simplify:

$$ \frac{x^2}{9}+\frac{y^2}{4}+\frac{4}{9}=1, \qquad \frac{x^2}{9}+\frac{y^2}{4}=\frac{5}{9}. $$

Divide by \(5/9\) to put the slice in standard ellipse form:

$$ \frac{x^2}{5}+\frac{y^2}{20/9}=1. $$

The semiaxes are \(\sqrt{5}\) and \(2\sqrt{5}/3\). Check using the original equation: at \(x=\sqrt{5},y=0,z=2\), the sum is \(5/9+0+4/9=1\); at \(x=0,y=2\sqrt{5}/3,z=2\), it is \(0+(20/9)/4+4/9=5/9+4/9=1\).

At \(z=3\), the equation becomes \(x^2/9+y^2/4+1=1\), so \(x^2/9+y^2/4=0\). Both terms are nonnegative, so \(x=y=0\): the slice is the single point \((0,0,3)\). Substitution gives \(0+0+9/9=1\), as required.

Cones: A Point and Two Expanding Directions

A double elliptic cone with vertex at the origin and axis along \(z\) can be written as

$$ \frac{x^2}{a^2}+\frac{y^2}{b^2}=\frac{z^2}{c^2}, \qquad a,b,c>0. $$

At height \(z=k\), the slice is

$$ \frac{x^2}{a^2}+\frac{y^2}{b^2}=\frac{k^2}{c^2}. $$

For \(k\ne 0\), this is an ellipse whose semiaxes are \(a|k|/c\) and \(b|k|/c\). At \(k=0\), it is just the vertex. The slices grow as \(|k|\) grows, both above and below the vertex. Unlike an ellipsoid, a double cone does not end at a largest height: its slices continue to expand as the distance from the vertex increases.

Worked Example: Measuring Cone Slices

Find the horizontal cross-sections at \(z=0\), \(z=2\), and \(z=-2\) for

$$ \frac{x^2}{4}+\frac{y^2}{9}=\frac{z^2}{16}. $$

At \(z=0\), the equation is \(x^2/4+y^2/9=0\). Each term is nonnegative, so \(x=y=0\). The slice is the vertex \((0,0,0)\), which satisfies the original equation.

At \(z=2\), substitute \(z=2\):

$$ \frac{x^2}{4}+\frac{y^2}{9}=\frac{4}{16}=\frac14. $$

Multiplying by \(4\) gives \(x^2+y^2/ (9/4)=1\), equivalently

$$ \frac{x^2}{1}+\frac{y^2}{9/4}=1. $$

The semiaxes are \(1\) and \(3/2\). Check at the \(x\)-endpoint \((1,0,2)\): the left side is \(1/4\), and the right side is \(4/16=1/4\). At the \(y\)-endpoint \((0,3/2,2)\), the left side is \((9/4)/9=1/4\), again equal to the right side.

At \(z=-2\), the squared height is again \(4\), so the cross-section has the same ellipse and the same semiaxes. The same endpoint checks hold because \(z^2=4\). This equality of slices at opposite heights reflects the cone's symmetry about the \(xy\)-plane.

Hyperboloids: Two Different Slice Patterns

A hyperboloid of one sheet with axis along \(z\) has the standard form

$$ \frac{x^2}{a^2}+\frac{y^2}{b^2}-\frac{z^2}{c^2}=1. $$

At height \(k\), its horizontal slice satisfies

$$ \frac{x^2}{a^2}+\frac{y^2}{b^2}=1+\frac{k^2}{c^2}. $$

The right-hand side is always positive. Every height has an elliptical cross-section. At \(k=0\), the ellipse has semiaxes \(a\) and \(b\); as \(|k|\) increases, the right-hand side increases and the slices grow. So the surface has a smallest horizontal slice at its middle, then widens in both directions.

A hyperboloid of two sheets with its axis along \(z\) is written

$$ \frac{z^2}{c^2}-\frac{x^2}{a^2}-\frac{y^2}{b^2}=1. $$

At height \(k\), rearranging gives

$$ \frac{x^2}{a^2}+\frac{y^2}{b^2}=\frac{k^2}{c^2}-1. $$

A real ellipse exists only when \(|k|>c\). At \(k=c\) or \(k=-c\), the section is a point, and for \(|k|<c\), it is empty. Beyond either boundary height, the ellipses grow as \(|k|\) increases. The empty band between the two sheets is the crucial feature distinguishing this surface from a one-sheet hyperboloid.

Worked Example: Finding the Allowed Slices of a Two-Sheet Hyperboloid

Analyze the horizontal sections of

$$ \frac{z^2}{4}-\frac{x^2}{9}-\frac{y^2}{4}=1 $$

at heights \(z=0\), \(z=2\), and \(z=3\). Rearrange the equation for a fixed height \(k\):

$$ \frac{x^2}{9}+\frac{y^2}{4}=\frac{k^2}{4}-1. $$

At \(k=0\), the right side is \(-1\). The left side is nonnegative, so there are no real points at that height.

At \(k=2\), the right side is \(4/4-1=0\). Therefore \(x=y=0\), giving the point \((0,0,2)\). Substitution into the original equation gives \(4/4-0-0=1\).

At \(k=3\), the right side is \(9/4-1=5/4\). Divide by \(5/4\):

$$ \frac{x^2}{45/4}+\frac{y^2}{5}=1. $$

The semiaxes are \(3\sqrt{5}/2\) and \(\sqrt{5}\). For a check, use the \(x\)-endpoint \(x=3\sqrt{5}/2,y=0,z=3\): the original left side is \(9/4-(45/4)/9-0=9/4-5/4=1\). At \(x=0,y=\sqrt{5},z=3\), it is \(9/4-0-5/4=1\). The ellipse is real, and its points satisfy the surface equation.

Why Substitution Gives the Cross-Section

The method is direct because a horizontal plane at height \(k\) consists exactly of the points whose \(z\)-coordinate equals \(k\). To find points belonging to both that plane and the surface, impose \(z=k\) in the surface equation. The remaining equation describes precisely the allowed \(x\)- and \(y\)-coordinates at that height.

For example, applying this reasoning to a one-sheet hyperboloid gives

$$ \frac{x^2}{a^2}+\frac{y^2}{b^2}-\frac{k^2}{c^2}=1. $$

Moving the fixed height term to the other side yields

$$ \frac{x^2}{a^2}+\frac{y^2}{b^2}=1+\frac{k^2}{c^2}. $$

The right-hand side is positive at every real height \(k\), so every horizontal plane intersects the surface. Since \(k^2\) increases with \(|k|\), the right side—and therefore the squared scale of the ellipse—increases as one moves away from the central slice. This derives both the existence of every slice and its growth pattern.

A Reliable Identification Procedure

When the equation or a sequence of slices is unfamiliar, separate the evidence into three questions: What shape is each nonempty section? At which heights does it exist? How does its size change with height?

1
Set the height.
Replace \(z\) by a fixed value \(k\) in the surface equation.
2
Check existence.
Read the right-hand side of the resulting ellipse equation. A positive value gives an ellipse, zero gives a point, and a negative value gives no real section.
3
Track the pattern.
Compare the ellipse scale as \(|k|\) changes, and note whether there is a finite range of allowed heights.

The patterns can be summarized as follows.

SurfaceHorizontal sectionsPattern with height
EllipsoidEllipses between two endpoint heights; points at the endpointsLargest at the central height, shrinking to points
Double elliptic coneEllipses except for the vertex pointExpands away from the vertex in both directions
Hyperboloid of one sheetAn ellipse at every heightSmallest at its central height, expanding in both directions
Hyperboloid of two sheetsPoints at two boundary heights, ellipses beyond them, empty betweenExpands outward from the two boundary points

A Common Sign Mistake

For a one-sheet hyperboloid, the negative \(z^2\) term must be moved carefully. Starting with

$$ \frac{x^2}{a^2}+\frac{y^2}{b^2}-\frac{z^2}{c^2}=1, $$

substituting \(z=k\) gives

$$ \frac{x^2}{a^2}+\frac{y^2}{b^2}-\frac{k^2}{c^2}=1. $$

A common incorrect step is to write the right side as \(1-k^2/c^2\). That is the wrong sign: adding \(k^2/c^2\) to both sides gives \(1+k^2/c^2\), not \(1-k^2/c^2\). The incorrect expression would imply that the sections disappear when \(|k|>c\), which contradicts the original equation. For instance, at \(k=2c\), the correct slice has right side \(1+4=5\), while the incorrect one gives \(-3\), an impossible ellipse. Substituting \(z=2c\) into the original equation confirms the correct relation: the \(x,y\) terms must sum to \(1+4=5\).

Key takeaway: To identify a surface from horizontal sections, track three features together: where sections exist, whether a boundary section is a point, and how the ellipse sizes change as the absolute height increases.

Check Your Understanding

Use substitution and the change in ellipse size to answer these questions.

  1. For an ellipsoid centered at the origin with \(z\)-semiaxis \(c\), at which heights is the horizontal section empty, a point, or an ellipse?
  2. A surface has elliptical sections at every height, with the smallest ellipse at \(z=0\) and larger ellipses as \(|z|\) increases. Which of the four surface patterns above does this match?
  3. For the cone \(x^2/4+y^2/9=z^2/16\), what are the semiaxis lengths of its section at \(z=4\)?
  4. A surface has no points for \(|z|<c\), points at \(z=\pm c\), and ellipses for \(|z|>c\). Identify its standard quadric type.
  5. For a one-sheet hyperboloid, explain why a negative \(z^2\) term leads to a plus sign on the right side after fixing the height.