Tutorials › Calculus III › Elliptic Cones and Their Vertex Cross-Sections

Home  /  Tutorials  /  Coordinates and Quadric Surfaces

Coordinates and Quadric Surfaces · Tutorial 14 of 999

Elliptic Cones and Their Vertex Cross-Sections

Use horizontal and vertical cross-sections to identify an elliptic cone, locate its vertex and axis, and distinguish its two nappes.

Beginner 10 min read

What You'll Learn

  • Recognize the standard equation of an elliptic cone with a coordinate-parallel axis
  • Locate the cone’s vertex and identify its axis and orientation
  • Derive horizontal cross-sections and determine their semi-axis lengths
  • Interpret the vertex section and vertical traces through the axis
  • Check an equation’s classification using points and cross-sections

How an Elliptic Cone Appears in an Equation

An elliptic cone can be recognized by a distinctive relationship between two squared transverse coordinates and the square of a third coordinate. In a typical position, the transverse coordinates describe an ellipse, while the third coordinate measures distance along the cone’s axis.

Definition: An elliptic cone with vertex \((h,k,\ell)\) and axis parallel to the \(z\)-axis has equation $$ \frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2} = \frac{(z-\ell)^2}{c^2}, \qquad a,b,c>0. $$ Its vertex is \((h,k,\ell)\), and its axis is the line \(x=h,\ y=k\). The cone extends in both directions along this line.

The positive constants \(a\), \(b\), and \(c\) determine the cone’s proportions. If \(a=b\), the horizontal cross-sections are circles and the cone is circular. If \(a\ne b\), those cross-sections are ellipses, so the cone is elliptic. The constant \(c\) controls how quickly the cross-sections widen as \(z\) moves away from \(\ell\).

The equation can also be written with the \(x\)- or \(y\)-axis as its axis. In each case, the squared coordinate along the axis appears by itself on one side of the equality, and the two squared transverse coordinates appear on the other. The corresponding axis line passes through the vertex and is parallel to that coordinate axis.

Horizontal Cross-Sections

A horizontal plane has equation \(z=t\), where \(t\) is a fixed number. Substituting \(z=t\) into the cone equation gives

$$ \frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2} = \frac{(t-\ell)^2}{c^2}. $$

If \(t\ne\ell\), divide both sides by \((t-\ell)^2/c^2\). The section is the ellipse

$$ \frac{(x-h)^2}{a^2(t-\ell)^2/c^2} + \frac{(y-k)^2}{b^2(t-\ell)^2/c^2} =1. $$

Its center is \((h,k,t)\), and its semi-axis lengths, along the \(x\)- and \(y\)-directions, are \(a|t-\ell|/c\) and \(b|t-\ell|/c\). Thus the ellipses grow in proportion to their distance from the vertex. At \(t=\ell\), both squared terms must be zero, so the horizontal section is just the vertex.

Key observation: Horizontal sections on opposite sides of the vertex have the same size when they are the same distance from it. The vertex section is a single point, and the nonvertex sections are ellipses centered on the axis.

Vertical Traces Through the Axis

A vertical plane containing the \(z\)-axis of the cone can be chosen by fixing a direction in the \(xy\)-plane. For example, the coordinate plane \(x=h\) contains the axis. Substitution into the cone equation gives

$$ \frac{(y-k)^2}{b^2}=\frac{(z-\ell)^2}{c^2}. $$

Taking both possible signs gives two lines:

$$ y-k=\frac{b}{c}(z-\ell) \qquad\text{or}\qquad y-k=-\frac{b}{c}(z-\ell). $$

These lines intersect at the vertex. The vertical trace in the plane \(y=k\) likewise consists of two intersecting lines. A vertical plane through the axis at another direction also cuts the cone in two lines, though their slopes in that plane depend on the chosen direction. The paired lines show how the two sides of the cone meet at the vertex.

Why These Cross-Sections Have These Shapes

The cross-section descriptions follow directly from substitution, but it is useful to see why the horizontal sections preserve the same shape as they move. Begin with a point on the cone and define scaled coordinates \(X=(x-h)/a\), \(Y=(y-k)/b\), and \(Z=(z-\ell)/c\). The equation becomes \(X^2+Y^2=Z^2\). At any fixed nonzero \(Z\), dividing by \(Z^2\) gives a circle of radius \(|Z|\) in the \(XY\)-coordinates. Returning to \(x\) and \(y\) scales the two directions by \(a\) and \(b\), producing an ellipse.

$$ X^2+Y^2=Z^2,\quad Z\ne0 \quad\Longrightarrow\quad \left(\frac{X}{|Z|}\right)^2+ \left(\frac{Y}{|Z|}\right)^2=1. $$

At \(Z=0\), the equation requires \(X^2+Y^2=0\). Since squares are nonnegative, this means \(X=Y=0\). That is why the section at the vertex is a point rather than an ellipse of positive size. The same equation also shows that if \((X,Y,Z)\) lies on the cone, then \((-X,-Y,-Z)\) lies on it. The surface therefore extends on both sides of the vertex.

Worked Example: Identify the Vertex and Read the Horizontal Sections

Worked Example: Identify the Vertex and Read the Horizontal Sections

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

Step 1: Identify the vertex and axis. Comparing with the standard form gives \(h=0,\ k=0,\ \ell=0\), so the vertex is \((0,0,0)\). The squared coordinate \(z\) is the axial coordinate, so the axis is the \(z\)-axis.

Step 2: Find the sections at \(z=4\) and \(z=-4\). Substituting \(z=4\) gives $$ \frac{x^2}{9}+\frac{y^2}{4}=\frac{16}{16}=1. $$ This ellipse has semi-axis lengths \(3\) and \(2\). Substituting \(z=-4\) gives the same right-hand side, since \((-4)^2/16=1\), and therefore the same ellipse.

Step 3: Check the vertex section. At \(z=0\), $$ \frac{x^2}{9}+\frac{y^2}{4}=0. $$ Both terms are nonnegative, so both must be zero. Thus \(x=0,\ y=0\), confirming that this section consists only of the vertex.

Check: The point \((3,0,4)\) satisfies the equation because \(3^2/9+0^2/4=1\) and \(4^2/16=1\). Its reflected point \((3,0,-4)\) also satisfies it because \((-4)^2/16=1\). The equation includes both sides of the cone.

Worked Example: A Shifted Cone with a Horizontal and a Vertical Trace

Worked Example: A Shifted Cone with a Horizontal and a Vertical Trace

Consider $$ \frac{(x-2)^2}{25}+\frac{(y+1)^2}{9}=\frac{(z-4)^2}{16}. $$

Step 1: Locate the vertex and axis. The equation is in standard form with \(h=2,\ k=-1,\ \ell=4\). The vertex is \((2,-1,4)\), and the axis is the line \(x=2,\ y=-1\), parallel to the \(z\)-axis.

Step 2: Take the horizontal section \(z=8\). Here \(z-\ell=8-4=4\), so $$ \frac{(x-2)^2}{25}+\frac{(y+1)^2}{9} =\frac{4^2}{16}=1. $$ The section is an ellipse centered at \((2,-1,8)\), with semi-axis lengths \(5\) and \(3\).

Step 3: Take the vertical section \(x=2\). Substituting \(x=2\) makes the first term zero: $$ \frac{(y+1)^2}{9}=\frac{(z-4)^2}{16}. $$ Taking square roots with both signs gives $$ \frac{y+1}{3}=\frac{z-4}{4} \quad\text{or}\quad \frac{y+1}{3}=-\frac{z-4}{4}. $$ Equivalently, \(y+1=3(z-4)/4\) or \(y+1=-3(z-4)/4\). These are two lines meeting at \((y,z)=(-1,4)\), the vertex in this plane.

Check: At the vertex, all three shifted coordinates are zero, so both sides of the original equation equal zero. The horizontal section at \(z=8\) also contains \((x,y)=(7,-1)\): the left side is \(5^2/25+0=1\), equal to the right side \(4^2/16=1\).

Worked Example: Recover a Cone from a Horizontal Section

Worked Example: Recover a Cone from a Horizontal Section

A cone has vertex \((0,0,0)\), axis along the \(z\)-axis, and the section at \(z=3\) is $$ \frac{x^2}{36}+\frac{y^2}{16}=1. $$ Find its equation, assuming its horizontal sections are centered on the axis and scale in proportion to their distance from the vertex.

Step 1: Use the standard form. Write the cone as $$ \frac{x^2}{a^2}+\frac{y^2}{b^2}=\frac{z^2}{c^2}. $$ At \(z=3\), the right-hand side is \(9/c^2\). Dividing the section equation by \(9/c^2\) shows that the coefficients of \(x^2\) and \(y^2\) in its unit-ellipse form have denominators \(9a^2/c^2\) and \(9b^2/c^2\). Matching these with \(36\) and \(16\) gives \(9a^2/c^2=36\) and \(9b^2/c^2=16\).

Step 2: Choose convenient proportional values. The equations imply \(a^2/c^2=4\) and \(b^2/c^2=16/9\). Taking \(c=3\) gives \(a=6\) and \(b=4\). Therefore one equation for the cone is $$ \frac{x^2}{36}+\frac{y^2}{16}=\frac{z^2}{9}. $$

Step 3: Check the given section and the vertex. At \(z=3\), the right-hand side is \(9/9=1\), reproducing the given ellipse. At \(z=0\), the equation requires \(x^2/36+y^2/16=0\), so \(x=y=0\). The section there is exactly the stated vertex.

Check the scaling: At \(z=6\), the right-hand side is \(36/9=4\). Dividing by \(4\) gives \(x^2/144+y^2/64=1\), so the semi-axis lengths are \(12\) and \(8\), twice those at \(z=3\), as required.

A Common Mistake: Keeping Only One Side of the Cone

Because the standard equation contains \(z^2\), solving for \(z\) requires both signs. For example, from $$ \frac{x^2}{9}+\frac{y^2}{4}=\frac{z^2}{16} $$ we obtain $$ z=\pm4\sqrt{\frac{x^2}{9}+\frac{y^2}{4}}. $$ Dropping the minus sign keeps only the part with \(z\ge0\), not the full cone. The original equation includes points with both positive and negative \(z\): for instance, \((3,0,4)\) and \((3,0,-4)\) both satisfy it. Always check whether solving for the axial coordinate has produced two signs.

Using Traces to Recognize the Surface

For an equation in the stated standard form, three checks help establish the geometry:

  • The vertex is where all three shifted coordinates are zero.
  • Fixing the axial coordinate away from the vertex produces an ellipse; fixing it at the vertex produces a point.
  • A vertical plane containing the axis produces two lines that meet at the vertex.

Together, these observations distinguish an elliptic cone from a surface whose horizontal sections stay the same size or whose vertical traces have another shape. They also give practical sketching information: mark the vertex and axis, draw a small ellipse near the vertex and a larger one farther away, and remember that matching heights on the two sides have matching cross-sections. The ellipses are cross-sections, not separate pieces of the surface; the cone connects them through the vertex.

Check Your Understanding

Use the equation and cross-section ideas above to answer these questions.

  1. For \(\frac{(x+1)^2}{4}+\frac{(y-3)^2}{25}=\frac{(z+2)^2}{9}\), give the vertex and axis.
  2. For that same cone, what are the semi-axis lengths of the horizontal section at \(z=1\)?
  3. What does the horizontal section at the vertex height look like, and why?
  4. Describe the vertical trace in the plane \(x=-1\) for the cone in question 1.
  5. Why does a squared axial coordinate mean that the equation includes points on both sides of the vertex?

Previous

Next

Back to Tutorials