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Study Guides > College Algebra: Co-requisite Course

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.[footnote]Architect of the Capitol. http://www.aoc.gov. Accessed April 15, 2014.[/footnote] It is an oval-shaped room called a whispering chamber because the shape makes it possible for sound to travel along the walls. 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.
A large ornate room with statues. Figure 1. The National Statuary Hall in Washington, D.C. (credit: Greg Palmer, Flickr)

Equations of Ellipses

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. Three images, each has two cones stacked on each other with the pointed ends joined in the middle. One pair of cones has a plane intersecting the bottom cone at an angle such that the plane passes through the sides of the cone. The outline of the intersection of the plane and the cone is an ellipse. The second pair of cones has a plane positioned vertically such that it is passing through the right hand side of the cones, creating hyperbolic shapes at the intersections of both the top and bottom cones. The third pair of cones has a plane intersecting the bottom cone such that the intersection of the plane and the cone is a parabolic shape. 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 [latex]\left(x,y\right)[/latex] 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. 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.

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 [latex]h[/latex] units horizontally and [latex]k[/latex] units vertically, the center of the ellipse will be [latex]\left(h,k\right)[/latex]. This translation results in the standard form of the equation we saw previously, with [latex]x[/latex] replaced by [latex]\left(x-h\right)[/latex] and y replaced by [latex]\left(y-k\right)[/latex].

A General Note: Standard Forms of the Equation of an Ellipse with Center (h, k)

The standard form of the equation of an ellipse with center [latex]\left(h,\text{ }k\right)[/latex] and major axis parallel to the x-axis is

[latex]\frac{{\left(x-h\right)}^{2}}{{a}^{2}}+\frac{{\left(y-k\right)}^{2}}{{b}^{2}}=1[/latex]

where
  • [latex]a>b[/latex]
  • the length of the major axis is [latex]2a[/latex]
  • the coordinates of the vertices are [latex]\left(h\pm a,k\right)[/latex]
  • the length of the minor axis is [latex]2b[/latex]
  • the coordinates of the co-vertices are [latex]\left(h,k\pm b\right)[/latex]
  • the coordinates of the foci are [latex]\left(h\pm c,k\right)[/latex], where [latex]{c}^{2}={a}^{2}-{b}^{2}[/latex].
The standard form of the equation of an ellipse with center [latex]\left(h,k\right)[/latex] and major axis parallel to the y-axis is

[latex]\frac{{\left(x-h\right)}^{2}}{{b}^{2}}+\frac{{\left(y-k\right)}^{2}}{{a}^{2}}=1[/latex]

where
  • [latex]a>b[/latex]
  • the length of the major axis is [latex]2a[/latex]
  • the coordinates of the vertices are [latex]\left(h,k\pm a\right)[/latex]
  • the length of the minor axis is [latex]2b[/latex]
  • the coordinates of the co-vertices are [latex]\left(h\pm b,k\right)[/latex]
  • the coordinates of the foci are [latex]\left(h,k\pm c\right)[/latex], where [latex]{c}^{2}={a}^{2}-{b}^{2}[/latex].
Just as with ellipses centered at the origin, ellipses that are centered at a point [latex]\left(h,k\right)[/latex] have vertices, co-vertices, and foci that are related by the equation [latex]{c}^{2}={a}^{2}-{b}^{2}[/latex]. We can use this relationship along with the midpoint and distance formulas to find the equation of the ellipse in standard form when the vertices and foci are given.
(a) Horizontal ellipse with center (h,k) (b) Vertical ellipse with center (h,k) (a) Horizontal ellipse with center [latex]\left(h,k\right)[/latex] (b) Vertical ellipse with center [latex]\left(h,k\right)[/latex]

How To: Given the vertices and foci of an ellipse not centered at the origin, write its equation in standard form.

  1. Determine whether the major axis is parallel to the x- or y-axis.
    1. If the y-coordinates of the given vertices and foci are the same, then the major axis is parallel to the x-axis. Use the standard form [latex]\frac{{\left(x-h\right)}^{2}}{{a}^{2}}+\frac{{\left(y-k\right)}^{2}}{{b}^{2}}=1[/latex].
    2. If the x-coordinates of the given vertices and foci are the same, then the major axis is parallel to the y-axis. Use the standard form [latex]\frac{{\left(x-h\right)}^{2}}{{b}^{2}}+\frac{{\left(y-k\right)}^{2}}{{a}^{2}}=1[/latex].
  2. Identify the center of the ellipse [latex]\left(h,k\right)[/latex] using the midpoint formula and the given coordinates for the vertices.
  3. Find [latex]{a}^{2}[/latex] by solving for the length of the major axis, [latex]2a[/latex], which is the distance between the given vertices.
  4. Find [latex]{c}^{2}[/latex] using [latex]h[/latex] and [latex]k[/latex], found in Step 2, along with the given coordinates for the foci.
  5. Solve for [latex]{b}^{2}[/latex] using the equation [latex]{c}^{2}={a}^{2}-{b}^{2}[/latex].
  6. Substitute the values for [latex]h,k,{a}^{2}[/latex], and [latex]{b}^{2}[/latex] into the standard form of the equation determined in Step 1.

Example: Writing the Equation of an Ellipse Centered at a Point Other Than the Origin

What is the standard form equation of the ellipse that has vertices [latex]\left(-2,-8\right)[/latex] and [latex]\left(-2,\text{2}\right)[/latex] and foci [latex]\left(-2,-7\right)[/latex] and [latex]\left(-2,\text{1}\right)?[/latex]

Answer: 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

[latex]\frac{{\left(x-h\right)}^{2}}{{b}^{2}}+\frac{{\left(y-k\right)}^{2}}{{a}^{2}}=1[/latex]

First, we identify the center, [latex]\left(h,k\right)[/latex]. The center is halfway between the vertices, [latex]\left(-2,-8\right)[/latex] and [latex]\left(-2,\text{2}\right)[/latex]. Applying the midpoint formula, we have:

[latex]\begin{array}{l}\left(h,k\right)=\left(\frac{-2+\left(-2\right)}{2},\frac{-8+2}{2}\right)\hfill \\ \text{ }=\left(-2,-3\right)\hfill \end{array}[/latex]

Next, we find [latex]{a}^{2}[/latex]. The length of the major axis, [latex]2a[/latex], is bounded by the vertices. We solve for [latex]a[/latex] by finding the distance between the y-coordinates of the vertices.

[latex]\begin{array}{c}2a=2-\left(-8\right)\\ 2a=10\\ a=5\end{array}[/latex]

So [latex]{a}^{2}=25[/latex]. Now we find [latex]{c}^{2}[/latex]. The foci are given by [latex]\left(h,k\pm c\right)[/latex]. So, [latex]\left(h,k-c\right)=\left(-2,-7\right)[/latex] and [latex]\left(h,k+c\right)=\left(-2,\text{1}\right)[/latex]. We substitute [latex]k=-3[/latex] using either of these points to solve for [latex]c[/latex].

[latex]\begin{array}{c}k+c=1\\ -3+c=1\\ c=4\end{array}[/latex] So [latex]{c}^{2}=16[/latex].

Next, we solve for [latex]{b}^{2}[/latex] using the equation [latex]{c}^{2}={a}^{2}-{b}^{2}[/latex].

[latex]\begin{array}{c}{c}^{2}={a}^{2}-{b}^{2}\\ 16=25-{b}^{2}\\ {b}^{2}=9\end{array}[/latex]

Finally, we substitute the values found for [latex]h,k,{a}^{2}[/latex], and [latex]{b}^{2}[/latex] into the standard form equation for an ellipse:

[latex]\frac{{\left(x+2\right)}^{2}}{9}+\frac{{\left(y+3\right)}^{2}}{25}=1[/latex]

Try It

What is the standard form equation of the ellipse that has vertices [latex]\left(-3,3\right)[/latex] and [latex]\left(5,3\right)[/latex] and foci [latex]\left(1 - 2\sqrt{3},3\right)[/latex] and [latex]\left(1+2\sqrt{3},3\right)?[/latex]

Answer: [latex]\frac{{\left(x - 1\right)}^{2}}{16}+\frac{{\left(y - 3\right)}^{2}}{4}=1[/latex]

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. 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.

Example: Locating the Foci of a Whispering Chamber

The Statuary Hall in the Capitol Building in Washington, D.C. is a whispering chamber. Its dimensions are 46 feet wide by 96 feet long. a. 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 [latex]\left(0,0\right)[/latex]. b. If two senators standing at the foci of this room can hear each other whisper, how far apart are the senators? Round to the nearest foot.

Answer: a. We are assuming a horizontal ellipse with center [latex]\left(0,0\right)[/latex], so we need to find an equation of the form [latex]\frac{{x}^{2}}{{a}^{2}}+\frac{{y}^{2}}{{b}^{2}}=1[/latex], where [latex]a>b[/latex]. We know that the length of the major axis, [latex]2a[/latex], is longer than the length of the minor axis, [latex]2b[/latex]. 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 [latex]a[/latex], we have [latex]2a=96[/latex], so [latex]a=48[/latex], and [latex]{a}^{2}=2304[/latex].
  • Solving for [latex]b[/latex], we have [latex]2b=46[/latex], so [latex]b=23[/latex], and [latex]{b}^{2}=529[/latex].
Therefore, the equation of the ellipse is [latex]\frac{{x}^{2}}{2304}+\frac{{y}^{2}}{529}=1[/latex]. b. To find the distance between the senators, we must find the distance between the foci, [latex]\left(\pm c,0\right)[/latex], where [latex]{c}^{2}={a}^{2}-{b}^{2}[/latex]. Solving for [latex]c[/latex], we have:

[latex]\begin{array}{ll}{c}^{2}={a}^{2}-{b}^{2}\hfill & \hfill \\ {c}^{2}=2304 - 529\hfill & \begin{array}{cccc}& & & \end{array}\text{Substitute using the values found in part (a)}.\hfill \\ c=\pm \sqrt{2304 - 529}\hfill & \begin{array}{cccc}& & & \end{array}\text{Take the square root of both sides}.\hfill \\ c=\pm \sqrt{1775} \hfill & \begin{array}{cccc}& & & \end{array}\text{Subtract}.\hfill \\ c\approx \pm 42\hfill & \begin{array}{cccc}& & & \end{array}\text{Round to the nearest foot}.\hfill \end{array}[/latex]

The points [latex]\left(\pm 42,0\right)[/latex] represent the foci. Thus, the distance between the senators is [latex]2\left(42\right)=84[/latex] feet.

Try It

Suppose a whispering chamber is 480 feet long and 320 feet wide. a. What is the standard form of the equation of the ellipse representing the room? Hint: assume a horizontal ellipse, and let the center of the room be the point [latex]\left(0,0\right)[/latex]. b. If two people are standing at the foci of this room and can hear each other whisper, how far apart are the people? Round to the nearest foot.

Answer: a. [latex]\frac{{x}^{2}}{57,600}+\frac{{y}^{2}}{25,600}=1[/latex] b. The people are standing 358 feet apart.

Graphs of Ellipses

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 [latex]\frac{{x}^{2}}{{a}^{2}}+\frac{{y}^{2}}{{b}^{2}}=1,\text{ }a>b[/latex] for horizontal ellipses and [latex]\frac{{x}^{2}}{{b}^{2}}+\frac{{y}^{2}}{{a}^{2}}=1,\text{ }a>b[/latex] for vertical ellipses.

How To: Given the standard form of an equation for an ellipse centered at [latex]\left(0,0\right)[/latex], sketch the graph.

  • Use the standard forms of the equations of an ellipse to determine the major axis, vertices, co-vertices, and foci.
    • If the equation is in the form [latex]\frac{{x}^{2}}{{a}^{2}}+\frac{{y}^{2}}{{b}^{2}}=1[/latex], where [latex]a>b[/latex], then
      • the major axis is the x-axis
      • the coordinates of the vertices are [latex]\left(\pm a,0\right)[/latex]
      • the coordinates of the co-vertices are [latex]\left(0,\pm b\right)[/latex]
      • the coordinates of the foci are [latex]\left(\pm c,0\right)[/latex]
    • If the equation is in the form [latex]\frac{{x}^{2}}{{b}^{2}}+\frac{{y}^{2}}{{a}^{2}}=1[/latex], where [latex]a>b[/latex], then
      • the major axis is the y-axis
      • the coordinates of the vertices are [latex]\left(0,\pm a\right)[/latex]
      • the coordinates of the co-vertices are [latex]\left(\pm b,0\right)[/latex]
      • the coordinates of the foci are [latex]\left(0,\pm c\right)[/latex]
  • Solve for [latex]c[/latex] using the equation [latex]{c}^{2}={a}^{2}-{b}^{2}[/latex].
  • Plot the center, vertices, co-vertices, and foci in the coordinate plane, and draw a smooth curve to form the ellipse.

Example: Graphing an Ellipse Centered at the Origin

Graph the ellipse given by the equation, [latex]\frac{{x}^{2}}{9}+\frac{{y}^{2}}{25}=1[/latex]. Identify and label the center, vertices, co-vertices, and foci.

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

  • the center of the ellipse is [latex]\left(0,0\right)[/latex]
  • the coordinates of the vertices are [latex]\left(0,\pm a\right)=\left(0,\pm \sqrt{25}\right)=\left(0,\pm 5\right)[/latex]
  • the coordinates of the co-vertices are [latex]\left(\pm b,0\right)=\left(\pm \sqrt{9},0\right)=\left(\pm 3,0\right)[/latex]
  • the coordinates of the foci are [latex]\left(0,\pm c\right)[/latex], where [latex]{c}^{2}={a}^{2}-{b}^{2}[/latex] Solving for [latex]c[/latex], we have:

[latex]\begin{array}{l}c=\pm \sqrt{{a}^{2}-{b}^{2}}\hfill \\ =\pm \sqrt{25 - 9}\hfill \\ =\pm \sqrt{16}\hfill \\ =\pm 4\hfill \end{array}[/latex]

Therefore, the coordinates of the foci are [latex]\left(0,\pm 4\right)[/latex]. Next, we plot and label the center, vertices, co-vertices, and foci, and draw a smooth curve to form the ellipse.

Try It

Graph the ellipse given by the equation [latex]\frac{{x}^{2}}{36}+\frac{{y}^{2}}{4}=1[/latex]. Identify and label the center, vertices, co-vertices, and foci.

Answer: center: [latex]\left(0,0\right)[/latex]; vertices: [latex]\left(\pm 6,0\right)[/latex]; co-vertices: [latex]\left(0,\pm 2\right)[/latex]; foci: [latex]\left(\pm 4\sqrt{2},0\right)[/latex]

Example: Graphing an Ellipse Centered at the Origin from an Equation Not in Standard Form

Graph the ellipse given by the equation [latex]4{x}^{2}+25{y}^{2}=100[/latex]. Rewrite the equation in standard form. Then identify and label the center, vertices, co-vertices, and foci.

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

[latex]\begin{array}{l} 4{x}^{2}+25{y}^{2}=100\hfill \\ \text{ }\frac{4{x}^{2}}{100}+\frac{25{y}^{2}}{100}=\frac{100}{100}\hfill \\ \text{ }\frac{{x}^{2}}{25}+\frac{{y}^{2}}{4}=1\hfill \end{array}[/latex]

Next, we determine the position of the major axis. Because [latex]25>4[/latex], the major axis is on the x-axis. Therefore, the equation is in the form [latex]\frac{{x}^{2}}{{a}^{2}}+\frac{{y}^{2}}{{b}^{2}}=1[/latex], where [latex]{a}^{2}=25[/latex] and [latex]{b}^{2}=4[/latex]. It follows that:
  • the center of the ellipse is [latex]\left(0,0\right)[/latex]
  • the coordinates of the vertices are [latex]\left(\pm a,0\right)=\left(\pm \sqrt{25},0\right)=\left(\pm 5,0\right)[/latex]
  • the coordinates of the co-vertices are [latex]\left(0,\pm b\right)=\left(0,\pm \sqrt{4}\right)=\left(0,\pm 2\right)[/latex]
  • the coordinates of the foci are [latex]\left(\pm c,0\right)[/latex], where [latex]{c}^{2}={a}^{2}-{b}^{2}[/latex]. Solving for [latex]c[/latex], we have:

[latex]\begin{array}{l}c=\pm \sqrt{{a}^{2}-{b}^{2}}\hfill \\ =\pm \sqrt{25 - 4}\hfill \\ =\pm \sqrt{21}\hfill \end{array}[/latex]

Therefore the coordinates of the foci are [latex]\left(\pm \sqrt{21},0\right)[/latex]. Next, we plot and label the center, vertices, co-vertices, and foci, and draw a smooth curve to form the ellipse.

Try It

Graph the ellipse given by the equation [latex]49{x}^{2}+16{y}^{2}=784[/latex]. Rewrite the equation in standard form. Then identify and label the center, vertices, co-vertices, and foci.

Answer: Standard form: [latex]\frac{{x}^{2}}{16}+\frac{{y}^{2}}{49}=1[/latex]; center: [latex]\left(0,0\right)[/latex]; vertices: [latex]\left(0,\pm 7\right)[/latex]; co-vertices: [latex]\left(\pm 4,0\right)[/latex]; foci: [latex]\left(0,\pm \sqrt{33}\right)[/latex]

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, [latex]\left(h,k\right)[/latex], we use the standard forms [latex]\frac{{\left(x-h\right)}^{2}}{{a}^{2}}+\frac{{\left(y-k\right)}^{2}}{{b}^{2}}=1,\text{ }a>b[/latex] for horizontal ellipses and [latex]\frac{{\left(x-h\right)}^{2}}{{b}^{2}}+\frac{{\left(y-k\right)}^{2}}{{a}^{2}}=1,\text{ }a>b[/latex] 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.

How To: Given the standard form of an equation for an ellipse centered at [latex]\left(h,k\right)[/latex], sketch the graph.

  • Use the standard forms of the equations of an ellipse to determine the center, position of the major axis, vertices, co-vertices, and foci.
    • If the equation is in the form [latex]\frac{{\left(x-h\right)}^{2}}{{a}^{2}}+\frac{{\left(y-k\right)}^{2}}{{b}^{2}}=1[/latex], where [latex]a>b[/latex], then
      • the center is [latex]\left(h,k\right)[/latex]
      • the major axis is parallel to the x-axis
      • the coordinates of the vertices are [latex]\left(h\pm a,k\right)[/latex]
      • the coordinates of the co-vertices are [latex]\left(h,k\pm b\right)[/latex]
      • the coordinates of the foci are [latex]\left(h\pm c,k\right)[/latex]
    • If the equation is in the form [latex]\frac{{\left(x-h\right)}^{2}}{{b}^{2}}+\frac{{\left(y-k\right)}^{2}}{{a}^{2}}=1[/latex], where [latex]a>b[/latex], then
      • the center is [latex]\left(h,k\right)[/latex]
      • the major axis is parallel to the y-axis
      • the coordinates of the vertices are [latex]\left(h,k\pm a\right)[/latex]
      • the coordinates of the co-vertices are [latex]\left(h\pm b,k\right)[/latex]
      • the coordinates of the foci are [latex]\left(h,k\pm c\right)[/latex]
  • Solve for [latex]c[/latex] using the equation [latex]{c}^{2}={a}^{2}-{b}^{2}[/latex].
  • Plot the center, vertices, co-vertices, and foci in the coordinate plane, and draw a smooth curve to form the ellipse.

Example: Graphing an Ellipse Centered at (h, k)

Graph the ellipse given by the equation, [latex]\frac{{\left(x+2\right)}^{2}}{4}+\frac{{\left(y - 5\right)}^{2}}{9}=1[/latex]. Identify and label the center, vertices, co-vertices, and foci.

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

  • the center of the ellipse is [latex]\left(h,k\right)=\left(-2,\text{5}\right)[/latex]
  • the coordinates of the vertices are [latex]\left(h,k\pm a\right)=\left(-2,5\pm \sqrt{9}\right)=\left(-2,5\pm 3\right)[/latex], or [latex]\left(-2,\text{2}\right)[/latex] and [latex]\left(-2,\text{8}\right)[/latex]
  • the coordinates of the co-vertices are [latex]\left(h\pm b,k\right)=\left(-2\pm \sqrt{4},5\right)=\left(-2\pm 2,5\right)[/latex], or [latex]\left(-4,5\right)[/latex] and [latex]\left(0,\text{5}\right)[/latex]
  • the coordinates of the foci are [latex]\left(h,k\pm c\right)[/latex], where [latex]{c}^{2}={a}^{2}-{b}^{2}[/latex]. Solving for [latex]c[/latex], we have:

[latex]\begin{array}{l}\begin{array}{l}\\ c=\pm \sqrt{{a}^{2}-{b}^{2}}\end{array}\hfill \\ =\pm \sqrt{9 - 4}\hfill \\ =\pm \sqrt{5}\hfill \end{array}[/latex]

Therefore, the coordinates of the foci are [latex]\left(-2,\text{5}-\sqrt{5}\right)[/latex] and [latex]\left(-2,\text{5+}\sqrt{5}\right)[/latex]. Next, we plot and label the center, vertices, co-vertices, and foci, and draw a smooth curve to form the ellipse.

Try It

Graph the ellipse given by the equation [latex]\frac{{\left(x - 4\right)}^{2}}{36}+\frac{{\left(y - 2\right)}^{2}}{20}=1[/latex]. Identify and label the center, vertices, co-vertices, and foci.

Answer: Center: [latex]\left(4,2\right)[/latex]; vertices: [latex]\left(-2,2\right)[/latex] and [latex]\left(10,2\right)[/latex]; co-vertices: [latex]\left(4,2 - 2\sqrt{5}\right)[/latex] and [latex]\left(4,2+2\sqrt{5}\right)[/latex]; foci: [latex]\left(0,2\right)[/latex] and [latex]\left(8,2\right)[/latex]

How To: Given the general form of an equation for an ellipse centered at (h, k), express the equation in standard form.

  • Recognize that an ellipse described by an equation in the form [latex]a{x}^{2}+b{y}^{2}+cx+dy+e=0[/latex] is in general form.
  • Rearrange the equation by grouping terms that contain the same variable. Move the constant term to the opposite side of the equation.
  • Factor out the coefficients of the [latex]{x}^{2}[/latex] and [latex]{y}^{2}[/latex] terms in preparation for completing the square.
  • Complete the square for each variable to rewrite the equation in the form of the sum of multiples of two binomials squared set equal to a constant, [latex]{m}_{1}{\left(x-h\right)}^{2}+{m}_{2}{\left(y-k\right)}^{2}={m}_{3}[/latex], where [latex]{m}_{1},{m}_{2}[/latex], and [latex]{m}_{3}[/latex] are constants.
  • Divide both sides of the equation by the constant term to express the equation in standard form.

Example: Graphing an Ellipse Centered at (h, k) by First Writing It in Standard Form

Graph the ellipse given by the equation [latex]4{x}^{2}+9{y}^{2}-40x+36y+100=0[/latex]. Identify and label the center, vertices, co-vertices, and foci.

Answer: We must begin by rewriting the equation in standard form. [latex-display]4{x}^{2}+9{y}^{2}-40x+36y+100=0[/latex-display] Group terms that contain the same variable, and move the constant to the opposite side of the equation. [latex-display]\left(4{x}^{2}-40x\right)+\left(9{y}^{2}+36y\right)=-100[/latex-display] Factor out the coefficients of the squared terms. [latex-display]4\left({x}^{2}-10x\right)+9\left({y}^{2}+4y\right)=-100[/latex-display] Complete the square twice. Remember to balance the equation by adding the same constants to each side. [latex-display]4\left({x}^{2}-10x+25\right)+9\left({y}^{2}+4y+4\right)=-100+100+36[/latex-display] Rewrite as perfect squares. [latex-display]4{\left(x - 5\right)}^{2}+9{\left(y+2\right)}^{2}=36[/latex-display] Divide both sides by the constant term to place the equation in standard form. [latex-display]\frac{{\left(x - 5\right)}^{2}}{9}+\frac{{\left(y+2\right)}^{2}}{4}=1[/latex-display] Now that the equation is in standard form, we can determine the position of the major axis. Because [latex]9>4[/latex], the major axis is parallel to the x-axis. Therefore, the equation is in the form [latex]\frac{{\left(x-h\right)}^{2}}{{a}^{2}}+\frac{{\left(y-k\right)}^{2}}{{b}^{2}}=1[/latex], where [latex]{a}^{2}=9[/latex] and [latex]{b}^{2}=4[/latex]. It follows that:

  • the center of the ellipse is [latex]\left(h,k\right)=\left(5,-2\right)[/latex]
  • the coordinates of the vertices are [latex]\left(h\pm a,k\right)=\left(5\pm \sqrt{9},-2\right)=\left(5\pm 3,-2\right)[/latex], or [latex]\left(2,-2\right)[/latex] and [latex]\left(8,-2\right)[/latex]
  • the coordinates of the co-vertices are [latex]\left(h,k\pm b\right)=\left(\text{5},-2\pm \sqrt{4}\right)=\left(\text{5},-2\pm 2\right)[/latex], or [latex]\left(5,-4\right)[/latex] and [latex]\left(5,\text{0}\right)[/latex]
  • the coordinates of the foci are [latex]\left(h\pm c,k\right)[/latex], where [latex]{c}^{2}={a}^{2}-{b}^{2}[/latex]. Solving for [latex]c[/latex], we have:
[latex-display]\begin{array}{l}c=\pm \sqrt{{a}^{2}-{b}^{2}}\hfill \\ =\pm \sqrt{9 - 4}\hfill \\ =\pm \sqrt{5}\hfill \end{array}[/latex-display] Therefore, the coordinates of the foci are [latex]\left(\text{5}-\sqrt{5},-2\right)[/latex] and [latex]\left(\text{5+}\sqrt{5},-2\right)[/latex]. Next we plot and label the center, vertices, co-vertices, and foci, and draw a smooth curve to form the ellipse.

Try It

Express the equation of the ellipse given in standard form. Identify the center, vertices, co-vertices, and foci of the ellipse. [latex-display]4{x}^{2}+{y}^{2}-24x+2y+21=0[/latex-display]

Answer: [latex-display]\frac{{\left(x - 3\right)}^{2}}{4}+\frac{{\left(y+1\right)}^{2}}{16}=1[/latex]; center: [latex]\left(3,-1\right)[/latex]; vertices: [latex]\left(3,-\text{5}\right)[/latex] and [latex]\left(3,\text{3}\right)[/latex]; co-vertices: [latex]\left(1,-1\right)[/latex] and [latex]\left(5,-1\right)[/latex]; foci: [latex]\left(3,-\text{1}-2\sqrt{3}\right)[/latex] and [latex]\left(3,-\text{1+}2\sqrt{3}\right)[/latex-display]

Key Equations

Horizontal ellipse, center at origin [latex]\frac{{x}^{2}}{{a}^{2}}+\frac{{y}^{2}}{{b}^{2}}=1,\text{ }a>b[/latex]
Vertical ellipse, center at origin [latex]\frac{{x}^{2}}{{b}^{2}}+\frac{{y}^{2}}{{a}^{2}}=1,\text{ }a>b[/latex]
Horizontal ellipse, center [latex]\left(h,k\right)[/latex] [latex]\frac{{\left(x-h\right)}^{2}}{{a}^{2}}+\frac{{\left(y-k\right)}^{2}}{{b}^{2}}=1,\text{ }a>b[/latex]
Vertical ellipse, center [latex]\left(h,k\right)[/latex] [latex]\frac{{\left(x-h\right)}^{2}}{{b}^{2}}+\frac{{\left(y-k\right)}^{2}}{{a}^{2}}=1,\text{ }a>b[/latex]

Key Concepts

  • An ellipse is the set of all points [latex]\left(x,y\right)[/latex] 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.

Glossary

center of an ellipse the midpoint of both the major and minor axes conic section any shape resulting from the intersection of a right circular cone with a plane ellipse the set of all points [latex]\left(x,y\right)[/latex] in a plane such that the sum of their distances from two fixed points is a constant foci plural of focus focus (of an ellipse) one of the two fixed points on the major axis of an ellipse such that the sum of the distances from these points to any point [latex]\left(x,y\right)[/latex] on the ellipse is a constant major axis the longer of the two axes of an ellipse minor axis the shorter of the two axes of an ellipse

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  • College Algebra. Provided by: OpenStax Authored by: Abramson, Jay et al.. License: CC BY: Attribution. License terms: Download for free at http://cnx.org/contents/[email protected].
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