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.
How to Read the Ellipse in a Slice
An equation of the form
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
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 \(z\):
At height \(z=k\), substitution gives
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
At \(z=0\), substitute \(0\) for \(z\):
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:
Divide by \(5/9\) to put the slice in standard ellipse form:
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
At height \(z=k\), the slice is
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
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 \(z=2\):
Multiplying by \(4\) gives \(x^2+y^2/ (9/4)=1\), equivalently
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
At height \(k\), its horizontal slice satisfies
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
At height \(k\), rearranging gives
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
at heights \(z=0\), \(z=2\), and \(z=3\). Rearrange the equation for a fixed height \(k\):
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 \(5/4\):
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
Moving the fixed height term to the other side yields
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?
Replace \(z\) by a fixed value \(k\) in the surface equation.
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.
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.
| Surface | Horizontal sections | Pattern with height |
|---|---|---|
| Ellipsoid | Ellipses between two endpoint heights; points at the endpoints | Largest at the central height, shrinking to points |
| Double elliptic cone | Ellipses except for the vertex point | Expands away from the vertex in both directions |
| Hyperboloid of one sheet | An ellipse at every height | Smallest at its central height, expanding in both directions |
| Hyperboloid of two sheets | Points at two boundary heights, ellipses beyond them, empty between | Expands 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
substituting \(z=k\) gives
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 \(1+4=5\).
Check Your Understanding
Use substitution and the change in ellipse size to answer these questions.
- For an ellipsoid centered at the origin with \(z\)-semiaxis \(c\), at which heights is the horizontal section empty, a point, or an ellipse?
- 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?
- For the cone \(x^2/4+y^2/9=z^2/16\), what are the semiaxis lengths of its section at \(z=4\)?
- A surface has no points for \(|z|<c\), points at \(z=\pm c\), and ellipses for \(|z|>c\). Identify its standard quadric type.
- 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.