Ellipsoids and Their Principal Cross-Sections

Recognizing an Ellipsoid

An ellipsoid is the three-dimensional counterpart of an ellipse. Its surface is closed and rounded in every direction, but its lengths along the three coordinate directions need not be equal. When the ellipsoid’s axes are parallel to the coordinate axes, its equation has a standard form that makes its center and dimensions visible.

To sketch one, read the equation carefully, mark its center and the endpoints of its principal axes, and use cross-sections to understand the surface between those endpoints. Coordinate-axis intercepts can also help, but a translated ellipsoid does not necessarily meet every coordinate axis.

Definition: An axis-aligned ellipsoid in standard form is the surface described by $$ \frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}+\frac{(z-\ell)^2}{c^2}=1, $$ where $a,b,c$ are positive real numbers. Its center is $(h,k,\ell)$, and its semiaxis lengths in the $x$-, $y$-, and $z$-directions are $a,b,c$, respectively.

The positive numbers $a,b,c$ are lengths, not squared lengths. Their squares appear as the denominators. If all three are equal, the equation describes a sphere; otherwise, the surface is stretched by different amounts in different coordinate directions. This tutorial concerns ellipsoids in the displayed standard form, whose principal directions are parallel to the coordinate axes.

Center, Semiaxes, and Endpoints

The center $(h,k,\ell)$ is the point at which each squared numerator is zero. To find the endpoints of the principal axes, vary one coordinate from the center while keeping the other two at their center values. For example, setting $y=k$ and $z=\ell$ leaves $(x-h)^2/a^2=1$, so $x=h\pm a$. The endpoints in the $x$-direction are therefore $(h-a,k,\ell)$ and $(h+a,k,\ell)$. The full length between them is $2a$, while $a$ is the semiaxis length.

The same reasoning gives the other four endpoints: $(h,k-b,\ell)$, $(h,k+b,\ell)$, $(h,k,\ell-c)$, and $(h,k,\ell+c)$. These six points are useful anchors for a sketch. They do not describe the whole surface, so the principal cross-sections provide additional information about its shape.

Principal Cross-Sections

The principal cross-sections are the traces made by the three coordinate-parallel planes through the center: $x=h$, $y=k$, and $z=\ell$. Substituting each plane equation into the ellipsoid equation removes one squared term and gives an ellipse.

\begin{aligned} x=h &: \quad \frac{(y-k)^2}{b^2}+\frac{(z-\ell)^2}{c^2}=1,\\ y=k &: \quad \frac{(x-h)^2}{a^2}+\frac{(z-\ell)^2}{c^2}=1,\\ z=\ell &: \quad \frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1. \end{aligned}

Each section is centered at the ellipsoid’s center within its plane. For instance, the section in $z=\ell$ has semiaxis lengths $a$ in the $x$-direction and $b$ in the $y$-direction. Thus the principal sections are not arbitrary additional curves: they show how pairs of the ellipsoid’s three semiaxes fit together.

Worked Example: A Centered Ellipsoid

Consider $$ \frac{x^2}{4}+\frac{y^2}{9}+z^2=1. $$ The center is $(0,0,0)$. Comparing denominators with the standard form gives $a^2=4$, $b^2=9$, and $c^2=1$, so the semiaxis lengths are $a=2$, $b=3$, and $c=1$. The longest direction is the $y$-direction and the shortest is the $z$-direction.

Substitute $x=0$, $y=0$, and $z=0$ in turn to find the principal sections: $$ \begin{aligned} x=0 &: \quad \frac{y^2}{9}+z^2=1,\\ y=0 &: \quad \frac{x^2}{4}+z^2=1,\\ z=0 &: \quad \frac{x^2}{4}+\frac{y^2}{9}=1. \end{aligned} $$ These are ellipses with semiaxis pairs $(3,1)$, $(2,1)$, and $(2,3)$, respectively.

For the coordinate-axis intercepts, set the other two coordinates to zero. On the $x$-axis, $y=z=0$, giving $x^2/4=1$, so $x=\pm2$. On the $y$-axis, $x=z=0$, giving $y^2/9=1$, so $y=\pm3$. On the $z$-axis, $x=y=0$, giving $z^2=1$, so $z=\pm1$.

Check each pair by substitution: $2^2/4=1$, $3^2/9=1$, and $1^2=1$. The negative endpoints give the same values because the coordinates are squared. A sketch should place all six intercepts and connect them with a closed surface consistent with the three elliptical principal sections.

Cross-Sections Away from the Center

Principal sections pass through the center, but a sketch can also be checked by examining a parallel plane away from the center. For example, in the plane $z=q$, the equation becomes $$ \frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2} =1-\frac{(q-\ell)^2}{c^2}. $$ The right-hand side determines what the section looks like. If it is positive, divide both sides by it to obtain an ellipse. If it is zero, the section consists only of the point $(h,k,q)$. If it is negative, no points satisfy the equation, so the plane does not meet the ellipsoid.

For a positive right-hand side, the section’s semiaxis lengths are $a$ and $b$, each multiplied by the square root of that right-hand side. As the plane moves away from $z=\ell$, the right-hand side decreases, and the elliptical section shrinks. This conclusion follows directly from the equation; it does not require a separate sketching rule.

Worked Example: A Section Away from the Center

For the ellipsoid $$ \frac{x^2}{4}+\frac{y^2}{9}+\frac{z^2}{16}=1, $$ find the cross-section in the plane $z=2$. Substituting $z=2$ gives $$ \frac{x^2}{4}+\frac{y^2}{9}+\frac{2^2}{16}=1. $$ Since $2^2/16=4/16=1/4$, this reduces to $$ \frac{x^2}{4}+\frac{y^2}{9}=\frac{3}{4}. $$ Divide each term by $3/4$, equivalently divide the entire equation by $3/4$: $$ \frac{x^2}{3}+\frac{y^2}{27/4}=1. $$ Thus the section is an ellipse centered at $(0,0,2)$, with semiaxis lengths $\sqrt{3}$ in the $x$-direction and $\sqrt{27/4}=3\sqrt{3}/2$ in the $y$-direction.

Check the two endpoints in the plane. At $x=\sqrt{3}, y=0, z=2$, the original left-hand side is $3/4+0+4/16=3/4+1/4=1$. At $x=0, y=3\sqrt{3}/2, z=2$, it is $0+(27/4)/9+4/16=3/4+1/4=1$. Both endpoints lie on the ellipsoid, as required.

The same equation also predicts what happens at the ends of the ellipsoid in the $z$-direction. At $z=4$, the equation requires $x^2/4+y^2/9=0$, which is possible only when $x=y=0$: the section is one point. At $z=5$, it would require $x^2/4+y^2/9=1-25/16=-9/16$, which is impossible because the left side is nonnegative. There is no section there.

Coordinate-Axis Intercepts

An intercept with a coordinate axis is a point on the ellipsoid that also lies on that axis. To test the $x$-axis, set $y=0$ and $z=0$; to test the $y$-axis, set $x=0$ and $z=0$; to test the $z$-axis, set $x=0$ and $y=0$. For a translated ellipsoid, these substitutions do not generally set the other squared terms to zero. Consequently, a coordinate axis may have two intersections, one intersection, or none. Do not assume that all three axes meet the surface.

Worked Example: A Translated Ellipsoid and Its Intercepts

Consider $$ \frac{(x-1)^2}{9}+\frac{(y+2)^2}{4}+\frac{(z-1)^2}{16}=1. $$ Its center is $(1,-2,1)$, and its semiaxis lengths are $3$, $2$, and $4$ along the $x$-, $y$-, and $z$-directions. The principal sections are found by fixing each coordinate at its center value: $$ \begin{aligned} x=1 &: \quad \frac{(y+2)^2}{4}+\frac{(z-1)^2}{16}=1,\\ y=-2 &: \quad \frac{(x-1)^2}{9}+\frac{(z-1)^2}{16}=1,\\ z=1 &: \quad \frac{(x-1)^2}{9}+\frac{(y+2)^2}{4}=1. \end{aligned} $$

On the $x$-axis, $y=0,z=0$. The two fixed terms are $$ \frac{(0+2)^2}{4}=1 \qquad\text{and}\qquad \frac{(0-1)^2}{16}=\frac{1}{16}. $$ They already sum to $17/16>1$, so adding the nonnegative $x$-term cannot give $1$. There are no $x$-axis intercepts. On the $z$-axis, $x=0,y=0$, and the fixed terms are $1/9$ and $1$, whose sum is $10/9>1$. There are no $z$-axis intercepts either.

On the $y$-axis, set $x=0,z=0$. The fixed terms sum to $$ \frac{(0-1)^2}{9}+\frac{(0-1)^2}{16} =\frac{1}{9}+\frac{1}{16} =\frac{16}{144}+\frac{9}{144} =\frac{25}{144}. $$ Therefore $$ \frac{(y+2)^2}{4}=1-\frac{25}{144}=\frac{119}{144}, \qquad (y+2)^2=\frac{119}{36}, \qquad y=-2\pm\frac{\sqrt{119}}{6}. $$ These are real $y$-axis intercepts. Substitution checks them: their squared $y$-term is $119/144$, and the other two terms are $16/144$ and $9/144$, so the total is $144/144=1$. A sketch must not invent $x$- or $z$-axis intercepts; the $y$-axis intersections alone are the coordinate-axis intercepts in this example.

A Reliable Sketching Procedure

The standard form gives a practical order for sketching. First, identify the center and mark it. Next, mark the two endpoints in each principal direction, using the semiaxis lengths as distances from the center. Then draw the three principal elliptical sections through the center. Finally, calculate any coordinate-axis intercepts that are useful and check them by substitution. The resulting surface should be closed, should pass through the marked endpoints, and should narrow to a point at each endpoint of a principal axis.

Read the standard form.
Match each squared numerator and denominator to the center coordinate and squared semiaxis length.

Mark the center and principal endpoints.
Move one semiaxis length in each positive and negative coordinate direction from the center.

Draw the principal sections.
Use the three ellipses in the planes through the center to guide the shape and relative widths.

Check intercepts and the surface.
Substitute coordinate-axis conditions, retain only real solutions, and verify any plotted point in the original equation.

Common Mistake: Reading the Denominator as a Length

A frequent error is to read a denominator as the semiaxis length instead of its square root. For example, in $$ \frac{x^2}{25}+\frac{y^2}{9}+\frac{z^2}{4}=1, $$ it is incorrect to say that the $x$-semiaxis has length $25$. The denominator is $25=a^2$, so $a=\sqrt{25}=5$. The $y$- and $z$-semiaxis lengths are $\sqrt{9}=3$ and $\sqrt{4}=2$, respectively.

The incorrect length $25$ would place an alleged endpoint at $(25,0,0)$. Substitution gives $25^2/25=25$, not $1$, so that point is not on the surface. The corrected endpoint $(5,0,0)$ gives $5^2/25=1$, which verifies the semiaxis length.

Why the Standard Form Determines a Closed Surface

The equation also explains why the graph is bounded. Each term is nonnegative, and their sum is $1$. Therefore no term can exceed $1$. In particular, $$ \frac{(x-h)^2}{a^2}\leq 1,\qquad \frac{(y-k)^2}{b^2}\leq 1,\qquad \frac{(z-\ell)^2}{c^2}\leq 1. $$ Taking square roots gives $|x-h|\leq a$, $|y-k|\leq b$, and $|z-\ell|\leq c$. Every point on the ellipsoid lies within these coordinate bounds. At the six extreme coordinate values, the other two terms must be zero, producing exactly the six principal endpoints. This is why those endpoints give reliable outer limits for a sketch.

Cross-sections and intercepts serve different purposes. The principal sections describe the ellipsoid through its center, while coordinate-axis intercepts depend on where the ellipsoid sits relative to the origin. For a translated ellipsoid, a missing coordinate-axis intersection does not mean the surface is missing an end; it means only that the particular axis does not pass through the surface.

Check Your Understanding

Use the standard form and show the substitutions that support your answers.

For $\frac{(x-2)^2}{16}+\frac{(y+1)^2}{4}+\frac{(z-3)^2}{25}=1$, state the center and all three semiaxis lengths.
Write the principal cross-section in the plane $z=3$ for the ellipsoid in question 1.
For $\frac{x^2}{9}+\frac{y^2}{4}+\frac{z^2}{16}=1$, describe the section in $z=5$: ellipse, point, or empty?
Find the coordinate-axis intercepts, if any, of $\frac{x^2}{4}+\frac{(y-1)^2}{4}+\frac{z^2}{9}=1$.
In $\frac{x^2}{36}+\frac{y^2}{49}+\frac{z^2}{4}=1$, what are the semiaxis lengths, and how can you check the positive $y$-endpoint?