The Ellipse

Can you imagine standing at one end of a large room and still being able to hear a whisper from a person standing at the other end? The National Statuary Hall in Washington, D.C., is such a room. It is a semi-circular room called a whispering chamber because the shape makes it possible for sound to travel along the walls and dome. In this section, we will investigate the shape of this room and its real-world applications, including how far apart two people in Statuary Hall can stand and still hear each other whisper.

Writing Equations of Ellipses in Standard Form

A conic section, or conic, is a shape resulting from intersecting a right circular cone with a plane. The angle at which the plane intersects the cone determines the shape, as shown below.

Conic sections can also be described by a set of points in the coordinate plane. Later in this chapter, we will see that the graph of any quadratic equation in two variables is a conic section. The signs of the equations and the coefficients of the variable terms determine the shape. This section focuses on the four variations of the standard form of the equation for the ellipse. An ellipse is the set of all points $(x,y)$ in a plane such that the sum of their distances from two fixed points is a constant. Each fixed point is called a focus (plural: foci).

We can draw an ellipse using a piece of cardboard, two thumbtacks, a pencil, and string. Place the thumbtacks in the cardboard to form the foci of the ellipse. Cut a piece of string longer than the distance between the two thumbtacks (the length of the string represents the constant in the definition). Tack each end of the string to the cardboard, and trace a curve with a pencil held taut against the string. The result is an ellipse.

Every ellipse has two axes of symmetry. The longer axis is called the major axis, and the shorter axis is called the minor axis. Each endpoint of the major axis is the vertex of the ellipse (plural: vertices), and each endpoint of the minor axis is a co-vertex of the ellipse. The center of an ellipse is the midpoint of both the major and minor axes. The axes are perpendicular at the center. The foci always lie on the major axis, and the sum of the distances from the foci to any point on the ellipse (the constant sum) is greater than the distance between the foci. See the figure below.

In this section, we restrict ellipses to those that are positioned vertically or horizontally in the coordinate plane. That is, the axes will either lie on or be parallel to the $x$- and $y$-axes. Later in the chapter, we will see ellipses that are rotated in the coordinate plane.

To work with horizontal and vertical ellipses in the coordinate plane, we consider two cases: those that are centered at the origin and those that are centered at a point other than the origin. First we will learn to derive the equations of ellipses, and then we will learn how to write the equations of ellipses in standard form. Later we will use what we learn to draw the graphs.

Deriving the Equation of an Ellipse Centered at the Origin

To derive the equation of an ellipse centered at the origin, we begin with the foci $(-c,0)$ and $(c,0)$. The ellipse is the set of all points $(x,y)$ such that the sum of the distances from $(x,y)$ to the foci is constant, as shown below.

If $(a,0)$ is a vertex of the ellipse, the distance from $(-c,0)$ to $(a,0)$ is $a-(-c)=a+c$. The distance from $(c,0)$ to $(a,0)$ is $a-c$. The sum of the distances from the foci to the vertex is

If $(x,y)$ is a point on the ellipse, then we can define the following variables:

By the definition of an ellipse, $d_1+d_2$ is constant for any point $(x,y)$ on the ellipse. We know that the sum of these distances is $2a$ for the vertex $(a,0)$. It follows that $d_1+d_2=2a$ for any point on the ellipse. We will begin the derivation by applying the distance formula. The rest of the derivation is algebraic.

Thus, the standard equation of an ellipse is $\tfrac{x^2}{a^2}+\tfrac{y^2}{b^2}=1$. This equation defines an ellipse centered at the origin. If $a>b$, the ellipse is stretched further in the horizontal direction, and if $b>a$, the ellipse is stretched further in the vertical direction.

Writing Equations of Ellipses Centered at the Origin in Standard Form

Standard forms of equations tell us about key features of graphs. Take a moment to recall some of the standard forms of equations we’ve worked with in the past: linear, quadratic, cubic, exponential, logarithmic, and so on. By learning to interpret standard forms of equations, we are bridging the relationship between algebraic and geometric representations of mathematical phenomena.

The key features of the ellipse are its center, vertices, co-vertices, foci, and lengths and positions of the major and minor axes. Just as with other equations, we can identify all of these features just by looking at the standard form of the equation. There are four variations of the standard form of the ellipse. These variations are categorized first by the location of the center (the origin or not the origin), and then by the position (horizontal or vertical). Each is presented along with a description of how the parts of the equation relate to the graph. Interpreting these parts allows us to form a mental picture of the ellipse.

(a) Horizontal ellipse with center $(0,0)$.

(b) Vertical ellipse with center $(0,0)$.

Example. What is the standard form equation of the ellipse that has vertices $(\pm8,0)$ and foci $(\pm5,0)$?

Solution. The foci are on the $x$-axis, so the major axis is the $x$-axis. Thus, the equation will have the form

The vertices are $(\pm8,0)$, so $a=8$ and $a^2=64$.

The foci are $(\pm5,0)$, so $c=5$ and $c^2=25$.

We know that the vertices and foci are related by the equation $c^2=a^2-b^2$. Solving for $b^2$, we have:

Now we need only substitute $a^2=64$ and $b^2=39$ into the standard form of the equation. The equation of the ellipse is $\tfrac{x^2}{64}+\tfrac{y^2}{39}=1$.

Writing Equations of Ellipses Not Centered at the Origin

Like the graphs of other equations, the graph of an ellipse can be translated. If an ellipse is translated $h$ units horizontally and $k$ units vertically, the center of the ellipse will be $(h,k)$. This translation results in the standard form of the equation we saw previously, with $x$ replaced by $(x-h)$ and $y$ replaced by $(y-k)$.

(a) Horizontal ellipse with center $(h,k)$.

(b) Vertical ellipse with center $(h,k)$.

Example. What is the standard form equation of the ellipse that has vertices $(-2,-8)$ and $(-2,2)$ and foci $(-2,-7)$ and $(-2,1)$?

Solution. The $x$-coordinates of the vertices and foci are the same, so the major axis is parallel to the $y$-axis. Thus, the equation of the ellipse will have the form

First, we identify the center, $(h,k)$. The center is halfway between the vertices, $(-2,-8)$ and $(-2,2)$. Applying the midpoint formula, we have:

Next, we find $a^2$. The length of the major axis, $2a$, is bounded by the vertices. We solve for $a$ by finding the distance between the $y$-coordinates of the vertices.

So $a^2=25$.

Now we find $c^2$. The foci are given by $(h,k\pm c)$. So, $(h,k-c)=(-2,-7)$ and $(h,k+c)=(-2,1)$. We substitute $k=-3$ using either of these points to solve for $c$.

So $c^2=16$.

Next, we solve for $b^2$ using the equation $c^2=a^2-b^2$.

Finally, we substitute the values found for $h,k,a^2,$ and $b^2$ into the standard form equation for an ellipse:

Graphing Ellipses Centered at the Origin

Just as we can write the equation for an ellipse given its graph, we can graph an ellipse given its equation. To graph ellipses centered at the origin, we use the standard form $\tfrac{x^2}{a^2}+\tfrac{y^2}{b^2}=1,\ a>b$ for horizontal ellipses and $\tfrac{x^2}{b^2}+\tfrac{y^2}{a^2}=1,\ a>b$ for vertical ellipses.

Example. Graph the ellipse given by the equation $\tfrac{x^2}{9}+\tfrac{y^2}{25}=1$. Identify and label the center, vertices, co-vertices, and foci.

Solution. First, we determine the position of the major axis. Because $25>9$, the major axis is on the $y$-axis. Therefore, the equation is in the form $\tfrac{x^2}{b^2}+\tfrac{y^2}{a^2}=1$, where $b^2=9$ and $a^2=25$. It follows that:

• the center of the ellipse is $(0,0)$
• the coordinates of the vertices are $(0,\pm a)=(0,\pm\sqrt{25})=(0,\pm5)$
• the coordinates of the co-vertices are $(\pm b,0)=(\pm\sqrt{9},0)=(\pm3,0)$
• the coordinates of the foci are $(0,\pm c)$, where $c^2=a^2-b^2$. Solving for $c$, we have:

Therefore, the coordinates of the foci are $(0,\pm4)$.

Next, we plot and label the center, vertices, co-vertices, and foci, and draw a smooth curve to form the ellipse.

Example. Graph the ellipse given by the equation $4x^2+25y^2=100$. Rewrite the equation in standard form. Then identify and label the center, vertices, co-vertices, and foci.

Solution. First, use algebra to rewrite the equation in standard form.

Next, we determine the position of the major axis. Because $25>4$, the major axis is on the $x$-axis. Therefore, the equation is in the form $\tfrac{x^2}{a^2}+\tfrac{y^2}{b^2}=1$, where $a^2=25$ and $b^2=4$. It follows that:

• the center of the ellipse is $(0,0)$
• the coordinates of the vertices are $(\pm a,0)=(\pm\sqrt{25},0)=(\pm5,0)$
• the coordinates of the co-vertices are $(0,\pm b)=(0,\pm\sqrt{4})=(0,\pm2)$
• the coordinates of the foci are $(\pm c,0)$, where $c^2=a^2-b^2$. Solving for $c$, we have:

Therefore the coordinates of the foci are $(\pm\sqrt{21},0)$.

Next, we plot and label the center, vertices, co-vertices, and foci, and draw a smooth curve to form the ellipse.

Graphing Ellipses Not Centered at the Origin

When an ellipse is not centered at the origin, we can still use the standard forms to find the key features of the graph. When the ellipse is centered at some point $(h,k)$, we use the standard forms $\tfrac{(x-h)^2}{a^2}+\tfrac{(y-k)^2}{b^2}=1,\ a>b$ for horizontal ellipses and $\tfrac{(x-h)^2}{b^2}+\tfrac{(y-k)^2}{a^2}=1,\ a>b$ for vertical ellipses. From these standard equations, we can easily determine the center, vertices, co-vertices, foci, and positions of the major and minor axes.

Example. Graph the ellipse given by the equation $\tfrac{(x+2)^2}{4}+\tfrac{(y-5)^2}{9}=1$. Identify and label the center, vertices, co-vertices, and foci.

Solution. First, we determine the position of the major axis. Because $9>4$, the major axis is parallel to the $y$-axis. Therefore, the equation is in the form $\tfrac{(x-h)^2}{b^2}+\tfrac{(y-k)^2}{a^2}=1$, where $b^2=4$ and $a^2=9$. It follows that:

• the center of the ellipse is $(h,k)=(-2,5)$
• the coordinates of the vertices are $(h,k\pm a)=(-2,5\pm\sqrt{9})=(-2,5\pm3)$, or $(-2,2)$ and $(-2,8)$
• the coordinates of the co-vertices are $(h\pm b,k)=(-2\pm\sqrt{4},5)=(-2\pm2,5)$, or $(-4,5)$ and $(0,5)$
• the coordinates of the foci are $(h,k\pm c)$, where $c^2=a^2-b^2$. Solving for $c$, we have:

Therefore, the coordinates of the foci are $(-2,5-\sqrt{5})$ and $(-2,5+\sqrt{5})$.

Next, we plot and label the center, vertices, co-vertices, and foci, and draw a smooth curve to form the ellipse.

Example. Graph the ellipse given by the equation $4x^2+9y^2-40x+36y+100=0$. Identify and label the center, vertices, co-vertices, and foci.

Solution. We must begin by rewriting the equation in standard form.

Group terms that contain the same variable, and move the constant to the opposite side of the equation.

Factor out the coefficients of the squared terms.

Complete the square twice. Remember to balance the equation by adding the same constants to each side.

Rewrite as perfect squares.

Divide both sides by the constant term to place the equation in standard form.

Now that the equation is in standard form, we can determine the position of the major axis. Because $9>4$, the major axis is parallel to the $x$-axis. Therefore, the equation is in the form $\tfrac{(x-h)^2}{a^2}+\tfrac{(y-k)^2}{b^2}=1$, where $a^2=9$ and $b^2=4$. It follows that:

• the center of the ellipse is $(h,k)=(5,-2)$
• the coordinates of the vertices are $(h\pm a,k)=(5\pm\sqrt{9},-2)=(5\pm3,-2)$, or $(2,-2)$ and $(8,-2)$
• the coordinates of the co-vertices are $(h,k\pm b)=(5,-2\pm\sqrt{4})=(5,-2\pm2)$, or $(5,-4)$ and $(5,0)$
• the coordinates of the foci are $(h\pm c,k)$, where $c^2=a^2-b^2$. Solving for $c$, we have:

Therefore, the coordinates of the foci are $(5-\sqrt{5},-2)$ and $(5+\sqrt{5},-2)$.

Next we plot and label the center, vertices, co-vertices, and foci, and draw a smooth curve to form the ellipse, as shown below.

Solving Applied Problems Involving Ellipses

Many real-world situations can be represented by ellipses, including orbits of planets, satellites, moons and comets, and shapes of boat keels, rudders, and some airplane wings. A medical device called a lithotripter uses elliptical reflectors to break up kidney stones by generating sound waves. Some buildings, called whispering chambers, are designed with elliptical domes so that a person whispering at one focus can easily be heard by someone standing at the other focus. This occurs because of the acoustic properties of an ellipse. When a sound wave originates at one focus of a whispering chamber, the sound wave will be reflected off the elliptical dome and back to the other focus, as shown below. In the whisper chamber at the Museum of Science and Industry in Chicago, two people standing at the foci — about 43 feet apart — can hear each other whisper. When these chambers are placed in unexpected places, such as the ones inside Bush International Airport in Houston and Grand Central Terminal in New York City, they can induce surprised reactions among travelers.

Example. A large room in an art gallery is a whispering chamber. Its dimensions are $46$ feet wide by $96$ feet long.

• What is the standard form of the equation of the ellipse representing the outline of the room? Hint: assume a horizontal ellipse, and let the center of the room be the point $(0,0)$.
• If two visitors standing at the foci of this room can hear each other whisper, how far apart are the two visitors? Round to the nearest foot.

Solution.

• We are assuming a horizontal ellipse with center $(0,0)$, so we need to find an equation of the form $\tfrac{x^2}{a^2}+\tfrac{y^2}{b^2}=1$, where $a>b$. We know that the length of the major axis, $2a$, is longer than the length of the minor axis, $2b$. So the length of the room, $96$, is represented by the major axis, and the width of the room, $46$, is represented by the minor axis.

  Solving for $a$, we have $2a=96$, so $a=48$, and $a^2=2{,}304$.

  Solving for $b$, we have $2b=46$, so $b=23$, and $b^2=529$.

  Therefore, the equation of the ellipse is $\tfrac{x^2}{2{,}304}+\tfrac{y^2}{529}=1$.

• To find the distance between the senators, we must find the distance between the foci, $(\pm c,0)$, where $c^2=a^2-b^2$. Solving for $c$, we have:

The points $(\pm42,0)$ represent the foci. Thus, the distance between the senators is $2(42)=84$ feet.

Key equations

Key concepts

• An ellipse is the set of all points $(x,y)$ in a plane such that the sum of their distances from two fixed points is a constant. Each fixed point is called a focus (plural: foci).
• When given the coordinates of the foci and vertices of an ellipse, we can write the equation of the ellipse in standard form.
• When given an equation for an ellipse centered at the origin in standard form, we can identify its vertices, co-vertices, foci, and the lengths and positions of the major and minor axes in order to graph the ellipse.
• When given the equation for an ellipse centered at some point other than the origin, we can identify its key features and graph the ellipse.
• Real-world situations can be modeled using the standard equations of ellipses and then evaluated to find key features, such as lengths of axes and distance between foci.

Practice

Write equations of ellipses in standard form

Graph ellipses centered at the origin

Graph ellipses not centered at the origin

Solve applied problems involving ellipses

This section is adapted from Precalculus 2e, Section 10.1: The Ellipse by Jay Abramson and OpenStax, © OpenStax, licensed under CC BY-NC-SA 4.0. Access the original for free at openstax.org. Changes: omitted a coreq-skills block the pinned CNXML prepends before the section proper (its own review of completing the square and graphing a circle, keyed to Intermediate Algebra 9.2.1 and 11.1.4, with its own worked examples and two “Practice Makes Perfect” exercise sets) — confirmed against the rendered PDF, page 1010 (true PDF index 1020), where the chapter-opening folio runs directly into the “10.1 The Ellipse” heading and its Learning Objectives with no corequisite-skills material between them. Omitted the National Statuary Hall photograph (a credit photograph, per house convention) but kept the paragraph that introduces it, describing the room generically rather than “as shown in Figure 1.” Recreated all twelve remaining source figures as accessible spec-first figures, comparing each against the rendered PDF: the three-cone conic-sections diagram (drawn as an exact schematic: each right double cone and its cutting plane, with the base circles, cone silhouettes, and the ellipse, hyperbola, and parabola traces all computed from the cone equation under a fixed oblique projection, hidden portions dashed) and the thumbtack-and-string construction (simplified to its essential geometry — an exactly sampled ellipse with the taut string drawn to the two foci; the pencil-and-hand illustration is represented by the string alone) as kind="figure" figures; the generic center/vertices/co-vertices/foci definition diagram, the $d_1$/$d_2$ derivation diagram, the two horizontal/vertical standard-forms figure pairs (center at the origin and at $(h,k)$), and all four worked-example answer graphs (Examples 3, 4, 5, and 6) as kind="graph" figures with the exact numbers each example derives; and the whispering-chamber reflective-property illustration and Example 7’s room diagram as kind="figure" ray diagrams. Every “write in standard form” fill-in declares answerForm="conic-standard-form", confirmed by replaying each printed general-form equation ($4x^2+40x+25y^2-100y+100=0$, $49x^2+16y^2=784$, $4x^2+y^2-24x+2y+21=0$) through the grader to confirm it reports form rather than correct. Every vertices/co-vertices/foci ask is keyed as a two-member answerMode="unordered" list of ordered pairs rather than a $\pm$ expression, per this chapter’s grading constraints. Six retained Try Its and Practice items ask for only one or two of a source exercise’s several requested quantities (e.g., “graph the ellipse, noting its center, vertices, and foci” is split into a standard-form or center fill-in plus a vertices or foci fill-in) rather than a single combined response, since the underlying quantities are not commensurable in one graded answer; each such split is disclosed here rather than merging unlike quantities into one component. Twelve selected end-of-section exercises (three per objective) were adapted into interactive Practice components, independently re-derived by hand and in Node rather than read off the source key; the applied Practice items’ rounded answers were checked against both full-precision and the source’s own rounded intermediates, and no discrepancy was found in this section.