15.1: The Ellipse

Learning Objectives 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.

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Figure \(\PageIndex{1}\)

The National Statuary Hall in Washington, D.C. (credit: Greg Palmer, Flickr)
 

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., shown in Figure \(\PageIndex{1}\), is such a room. 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.

Ellipses as Conic Sections

We begin this material with the general concept of a conic section.

Definition: Conic Section

A conic section, also known simply as a conic, is a curve formed by the intersection of a plane and a double-napped right circular cone.

The angle at which the plane intersects the cone determines the shape, as shown in Figure \(\PageIndex{2}\).

Figure \(\PageIndex{2}\)
 

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 within 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.

Definition: Ellipse

An ellipse is the set of all points 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. See Figure \(\PageIndex{3}\).

Figure \(\PageIndex{3}\)
 

Every ellipse has two axes of symmetry and several key points.

Definition: Major and Minor Axes, Vertex and Co-Vertex, and Center of an Ellipse

The longer axis of an ellipse is called the major axis, and the shorter axis is called the minor axis. Each endpoint of the major axis is called a vertex of the ellipse (plural: vertices), and each endpoint of the minor axis is called a co-vertex of the ellipse. The center of the 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 Figure \(\PageIndex{4}\).

Figure \(\PageIndex{4}\)
 

In this course, 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.

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.

Standard Form of an Ellipse Centered at the Origin

Since there are two ways we are going to orient our ellipses (horizontally and vertically), there are two equations for the standard form of an ellipse.

Theorem: Standard Form of the Equation of an Ellipse Centered at the Origin

The standard form of the equation of an ellipse with center \((0,0)\) and major axis on the \(x\)-axis is\[\dfrac{x^2}{v_x^2}+\dfrac{y^2}{v_y^2}=1, \nonumber \]where

• \( v_x \) and \( v_y \) are positive
• \(v_x > v_y\)
• the length of the major axis is \(2 v_x\)
• the coordinates of the vertices are \(( \pm v_x, 0)\)
• the length of the minor axis is \(2 v_y\)
• the coordinates of the co-vertices are \((0, \pm v_y)\)
• the coordinates of the foci are \(( \pm c, 0)\), where \(c^2= \left| v_x^2 - v_y^2 \right| \).

The standard form of the equation of an ellipse with center \((0,0)\) and major axis on the \(y\)-axis is\[ \dfrac{x^2}{v_x^2}+\dfrac{y^2}{v_y^2}=1, \nonumber \]where

• \( v_x \) and \( v_y \) are positive
• \(v_y > v_x \)
• the length of the major axis is \(2 v_y\)
• the coordinates of the vertices are \((0, \pm v_y)\)
• the length of the minor axis is \(2 v_x\)
• the coordinates of the co-vertices are \(( \pm v_x, 0)\)
• the coordinates of the foci are \((0, \pm c)\), where \(c^2 = \left| v_x^2 - v_y^2 \right| \).

Proof

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 in the figure below.

If \((v_x, 0)\) is a vertex of the ellipse (on the major axis), the distance from \((-c, 0)\) to \((v_x, 0)\) is \(v_x-(-c)=v_x+c\). The distance from \((c, 0)\) to \((v_x, 0)\) is \(v_x-c\). The sum of the distances from the foci to the vertex is\[(v_x+c)+(v_x-c) = 2v_x \nonumber \]If \((x, y)\) is a point on the ellipse, then we can define the following variables:\[ \begin{array}{rcl} d_1 & = & \text { the distance from }(-c, 0) \text { to }(x, y) \\[6pt] d_2 & = & \text { the distance from }(c, 0) \text { to }(x, y)\\[6pt] \end{array} \nonumber \]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 \(2 v_x\) for the vertex \((v_x, 0)\). It follows that \(d_1+d_2=2 v_x\) for any point on the ellipse. We will begin the derivation by applying the Distance Formula. The rest of the derivation is algebraic.\[ \begin{array}{rrclcl}
& d_1 + d_2 & = & 2v_x & & \\[6pt]
\implies & \sqrt{\left( x - (-c) \right)^2 + \left( y - 0 \right)^2} + \sqrt{\left( x - c \right)^2 + \left( y - 0 \right)^2} & = & 2v_x & \quad & \left( \text{substituting} \right) \\[6pt]
\implies & \sqrt{\left( x + c \right)^2 + y^2} + \sqrt{\left( x - c \right)^2 + y^2} & = & 2v_x & \quad & \left( \text{simplifying} \right) \\[6pt]
\implies & \sqrt{\left( x + c \right)^2 + y^2} & = & 2v_x - \sqrt{\left( x - c \right)^2 + y^2} & \quad & \left( \text{isolating a radical} \right) \\[6pt]
\implies & \left( x + c \right)^2 + y^2 & = & \left[ 2v_x - \sqrt{\left( x - c \right)^2 + y^2} \right]^2 & \quad & \left( \text{squaring both sides} \right) \\[6pt]
\implies & \left( x + c \right)^2 + y^2 & = & 4v_x^2 - 4v_x \sqrt{\left( x - c \right)^2 + y^2} + \left( x - c \right)^2 + y^2 & \quad & \left( \text{distributing} \right) \\[6pt]
\implies & x^2 + 2 c x + c^2 + y^2 & = & 4v_x^2 - 4v_x \sqrt{\left( x - c \right)^2 + y^2} + x^2 - 2c x + c^2 + y^2 & \quad & \left( \text{distributing} \right) \\[6pt]
\implies & 2 c x & = & 4v_x^2 - 4v_x \sqrt{\left( x - c \right)^2 + y^2} - 2c x & \quad & \left( \text{subtracting }x^2, \, c^2, \text{ and }y^2 \right. \\[6pt]
 &  &  & & \quad & \left. \text{ from both sides} \right) \\[6pt]
\implies & 4 c x - 4v_x^2 & = & -4v_x \sqrt{\left( x - c \right)^2 + y^2} & \quad & \left( \text{isolating the radical} \right) \\[6pt]
\implies & c x - v_x^2 & = & -v_x \sqrt{\left( x - c \right)^2 + y^2} & \quad & \left( \text{dividing both sides by }4 \right) \\[6pt]
\implies & \left(c x - v_x^2\right)^2 & = & v_x^2 \left(\left( x - c \right)^2 + y^2\right) & \quad & \left( \text{squaring both sides} \right) \\[6pt]
\implies & c^2 x^2 - 2v_x^2 c x + v_x^4 & = & v_x^2 \left(x^2 - 2cx + c^2 + y^2\right) & \quad & \left( \text{distributing} \right) \\[6pt]
\implies & c^2 x^2 - 2v_x^2 c x + v_x^4 & = & v_x^2 x^2 - 2 v_x^2 cx + v_x^2 c^2 + v_x^2 y^2 & \quad & \left( \text{distributing} \right) \\[6pt]
\implies & c^2 x^2 + v_x^4 & = & v_x^2 x^2 + v_x^2 c^2 + v_x^2 y^2 & \quad & \left( \text{adding }2v_x^2 c x\text{ to both sides} \right) \\[6pt]
\implies & v_x^4 - v_x^2 c^2 & = & v_x^2 x^2 - c^2 x^2 + v_x^2 y^2 & \quad & \left( \text{subtracting }v_x^2 c^2 \text{ and }c^2 x^2 \right. \\[6pt]
 &  &  & & \quad & \left. \text{ from both sides} \right) \\[6pt]
\implies & v_x^2(v_x^2 - c^2) & = & x^2(v_x^2 - c^2 ) + v_x^2 y^2 & \quad & \left( \text{factor} \right) \\[6pt]
\implies & v_x^2 b^2 & = & x^2 b^2 + v_x^2 y^2 & \quad & \left( \text{let }b^2 = v_x^2 - c^2 \right) \\[6pt]
\implies & \dfrac{v_x^2 b^2}{v_x^2 b^2} & = & \dfrac{x^2 b^2}{v_x^2 b^2} + \dfrac{v_x^2 y^2}{v_x^2 b^2} & \quad & \left( \text{divide both sides by }v_x^2 b^2 \right) \\[6pt]
\implies & 1 & = & \dfrac{x^2}{v_x^2} + \dfrac{y^2}{b^2} & \quad & \left( \text{simplify} \right) \\[6pt]
\end{array} \nonumber \]If we let \( x = 0 \) in our final result, we get \( y^2 = b^2 \), which implies \( y = \pm b \). Therefore, the point \( \left( 0,b \right) \) must be the co-vertex of the ellipse. As such, we relabel \( b = v_y \) to arrive at the standard equation of an ellipse:\[\dfrac{x^2}{v_x^2}+\dfrac{y^2}{v_y^2}=1. \nonumber \]This equation defines an ellipse centered at the origin. If \( v_x > v_y\), the ellipse is stretched further in the horizontal direction, and if \(v_y>v_x\), the ellipse is stretched further in the vertical direction.

You should spend time reading through the proof of this theorem (or, better yet, you should ask your instructor to go over this theorem in class!).

This theorem can be shortened quite a bit by saying that the larger of \( v_x \) and \( v_y \) dictates the major axis (and the smaller dictates the minor axis). In any case, the vertices in the horizontal direction are located \( v_x \) units from the center (at \( \left( \pm v_x,0 \right) \)) and the vertices in the vertical direction are \( v_y \) units from the center (at \( \left( 0, \pm v_y \right) \)). The foci are located \( c \) units from the center along the major axis, where \( c^2 = \left| v_x^2 - v_y^2 \right| \). The absolute values in this final equation force positivity of the right side.

Example \(\PageIndex{1}\): Writing the Equation of an Ellipse Centered at the Origin in Standard Form

What is the standard form equation of the ellipse that has vertices \(( \pm 8,0)\) and foci \(( \pm 5,0)\) ?

Solution

The foci are on the \(x\)-axis, so the major axis is the \(x\)-axis. Thus, the equation will have the form\[\dfrac{x^2}{v_x^2}+\dfrac{y^2}{v_y^2}=1 \nonumber \]The vertices are \(( \pm 8,0)\), so \(v_x=8\) and \(v_x^2=64\).

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

We know that the vertices and foci are related by the equation \(c^2 = \left| v_x^2 - v_y^2 \right| \). Solving for \(v_y^2\), we have:\[ \begin{array}{rrclcl} & c^2 & = & \left| v_x^2 - v_y^2 \right| & & \\[6pt] \implies & 25 & = & \left| 64 - v_y^2 \right| & \quad & \left( \text{substituting} \right) \\[6pt] \implies & 25 & = & 64 - v_y^2 & \quad & \left( \text{since the major axis is horizontal, }v_x^2 > v_y^2 \right) \\[6pt] \implies & v_y^2 & = & 39 & \quad & \left( \text{solving for }v_y^2 \right) \\[6pt] \end{array} \nonumber \]Now we need only substitute \(v_x^2=64\) and \(v_y^2=39\) into the standard form of the equation. The equation of the ellipse is \(\frac{x^2}{64}+\frac{y^2}{39}=1\).

Checkpoint \( \PageIndex{ 1 } \)

What is the standard form equation of the ellipse that has vertices \((0, \pm 4)\) and foci \((0, \pm \sqrt{15})\) ?

Q&A

Can we write the equation of an ellipse centered at the origin given coordinates of just one focus and vertex?

Yes. Ellipses are symmetrical, so the coordinates of the vertices of an ellipse centered around the origin will always have the form \(( \pm v_x, 0)\) or \((0, \pm v_y)\). Similarly, the coordinates of the foci will always have the form \(( \pm c, 0)\) or \((0, \pm c)\). Knowing this, we can use \(v_x\) (or \( v_y \)) and \(c\) from the given points, along with the equation \(c^2 = \left| v_x^2 - v_y^2 \right| \), to find the missing denominator.

Standard Form of an Ellipse Centered Off the Origin

Like the graphs of other equations, the graph of an ellipse can be transformed - specifically, in this conversation, shifted. If an ellipse is shifted \( h \) units horizontally and \(k\) units vertically, the center of the ellipse will be \((h,k)\). This transformation results in the standard form of the equation we saw previously, with \(x\) replaced by \((x−h)\) and \(y\) replaced by \((y−k)\).

Theorem: The Standard Form of the Equation of an Ellipse with Center \((h, k)\)

The standard form of the equation of an ellipse with center \((h, k)\) and major axis parallel to the \(x\)-axis is\[\dfrac{(x-h)^2}{v_x^2}+\dfrac{(y-k)^2}{v_y^2}=1, \nonumber \]where

• \( v_x \) and \( v_y \) are positive
• \(v_x^2 > v_y^2 \)
• the length of the major axis is \(2 v_x\)
• the coordinates of the vertices are \((h \pm v_x, k)\)
• the length of the minor axis is \(2 v_y\)
• the coordinates of the co-vertices are \((h, k \pm v_y)\)
• the coordinates of the foci are \((h \pm c, k)\), where \(c^2 = \left| v_x^2 - v_y^2 \right| \).

The standard form of the equation of an ellipse with center \((h, k)\) and major axis parallel to the \(y\)-axis is\[\dfrac{(x-h)^2}{v_x^2}+\dfrac{(y-k)^2}{v_y^2}=1, \nonumber \]where

• \( v_x \) and \( v_y \) are positive
• \(v_y^2 > v_x^2 \)
• the length of the major axis is \(2 v_y\)
• the coordinates of the vertices are \((h, k \pm v_y)\)
• the length of the minor axis is \(2 v_x\)
• the coordinates of the co-vertices are \((h \pm v_x, k)\)
• the coordinates of the foci are \((h, k \pm c)\), where \(c^2 = \left| v_x^2 - v_y^2 \right| \).

Again, this theorem can be shortened to saying that the larger of \( v_x \) and \( v_y \) dictates the major axis. In any case, the vertices in the horizontal direction are located \( v_x \) units from the center (at \( \left( h \pm v_x,k \right) \)) and the vertices in the vertical direction are \( v_y \) units from the center (at \( \left( h, k \pm v_y \right) \)). The foci are located \( c \) units from the center along the major axis, where \( c^2 = \left| v_x^2 - v_y^2 \right| \).

Example \(\PageIndex{2}\): 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 \((-2,-8)\) 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\[ \dfrac{(x-h)^2}{v_x^2}+\dfrac{(y-k)^2}{v_y^2}=1 \nonumber \]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:\[ \begin{array}{rcl} (h, k) & = & \left(\dfrac{-2+(-2)}{2}, \dfrac{-8+2}{2}\right) \\[6pt] & = & (-2,-3) \\[6pt] \end{array} \nonumber \]Next, we find \(v_y^2\). The length of the major axis, \(2 v_y\), is bounded by the vertices. We solve for \(v_y\) by finding the distance between the \(y\)-coordinates of the vertices.\[ \begin{array}{rrcl} & 2 v_y & = & 2-(-8) \\[6pt] \implies & 2 v_y & = & 10 \\[6pt] \implies & v_y & = & 5 \\[6pt] \end{array} \nonumber \]So \(v_y^2=25\).

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 \(c\).\[ \begin{array}{rrcl} & k+c & = & 1 \\[6pt] \implies & -3+c & = & 1 \\[6pt] \implies & c & = & 4 \\[6pt] \end{array} \nonumber \]So \(c^2=16\).

Next, we solve for \(v_x^2\) using the equation \(c^2 = \left| v_x^2 - v_y^2 \right| \).\[ \begin{array}{rrclcl} & c^2 & = & \left| v_x^2 - v_y^2 \right| & & \\[6pt] \implies & 16 & = & \left| v_x^2 - 25 \right| & \quad & \left( \text{substituting} \right) \\[6pt] \implies & 16 & = & 25 - v_x^2 & \quad & \left( \text{since }v_x^2 > v_x^2 \right) \\[6pt] \implies & v_x^2 & = & 9 & \quad & \left( \text{solving for }v_x^2 \right) \\[6pt] \end{array} \nonumber \]Finally, we substitute the values found for \(h\), \(k\), \(v_x^2\), and \(v_y^2\) into the standard form equation for an ellipse:\[ \dfrac{(x+2)^2}{9}+\dfrac{(y+3)^2}{25}=1 \nonumber \]

Checkpoint \( \PageIndex{ 2 } \)

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

Graphing Ellipses Centered at the Origin

Just as we can write the equation for an ellipse given information about its graph, we can graph an ellipse given its equation. To graph ellipses centered at the origin, we use the standard form \(\frac{x^2}{v_x^2}+\frac{y^2}{v_y^2}=1, a>b\).

Example \(\PageIndex{3}\): Graphing an Ellipse Given an Equation

Graph the ellipse given by the equation, \(\frac{x^2}{9}+\frac{y^2}{25}=1\). Identify and label the center, vertices, covertices, and foci.

Solution

From the equation, \(v_x^2=9\) and \(v_y^2=25\). Moreover, \( h = 0 \) and \( k = 0 \). It follows that:

• the major axis is vertical (\(v_y^2 > v_x^2\))
• the center of the ellipse is \((0,0)\)
• the coordinates of the vertices are \((0, \pm v_y)=(0, \pm \sqrt{25})=(0, \pm 5)\)
• the coordinates of the co-vertices are \(( \pm v_x, 0)=( \pm \sqrt{9}, 0)=( \pm 3,0)\)
• the coordinates of the foci are \((0, \pm c)\), where \(c^2 = \left| v_x^2 - v_y^2 \right| \). Solving for \(c\), we have:\[ \begin{array}{rrcl} & c^2 & = & \left| 9 - 25 \right| \\[6pt] & & = & \left| -16 \right| \\[6pt] & & = & 16 \\[6pt] \implies & c & = & 4 \\[6pt] \end{array} \nonumber \]

Therefore, the coordinates of the foci are \((0, \pm 4)\).

Next, we plot and label the center, vertices, co-vertices, and foci, and draw a smooth curve to form the ellipse. See Figure \(\PageIndex{5}\)

Figure \(\PageIndex{5}\)
 

Checkpoint \( \PageIndex{ 3 } \)

Graph the ellipse given by the equation \(\frac{x^2}{36}+\frac{y^2}{4}=1\). Identify and label the center, vertices, covertices, and foci.

Exercise \(\PageIndex{4}\): Graphing an Ellipse Centered at the Origin from an Equation Not in Standard Form

Graph the ellipse given by the equation \(4 x^2+25 y^2=100\). 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.\[ \begin{array}{rrcl} & 4 x^2+25 y^2 & = & 100 \\[6pt] \implies & \dfrac{4 x^2}{100}+\dfrac{25 y^2}{100} & = & \dfrac{100}{100} \\[6pt] \implies & \dfrac{x^2}{25}+\dfrac{y^2}{4} & = & 1 \\[6pt] \end{array}\nonumber \]From the equation, \(v_x^2=25\) and \(v_y^2=4\). Moreover, \( h = 0 \) and \( k = 0 \). It follows that:

• the major axis is horizontal (\(v_x^2 > v_y^2\))
• the center of the ellipse is \((0,0)\)
• the coordinates of the vertices are \((\pm v_x, 0)=(\pm \sqrt{25}, 0)=(\pm 5, 0)\)
• the coordinates of the co-vertices are \((0, \pm v_y)=(0, \pm \sqrt{4})=(0, \pm 2)\)
• the coordinates of the foci are \((\pm c,0)\), where \(c^2 = \left| v_x^2 - v_y^2 \right| \). Solving for \(c\), we have:\[ \begin{array}{rrcl} & c^2 & = & \left| 25 - 4 \right| \\[6pt] & & = & \left| 21 \right| \\[6pt] & & = & 21 \\[6pt] \implies & c & = & \sqrt{21} \\[6pt] \end{array} \nonumber \]

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.

Figure \(\PageIndex{6}\)
 

Checkpoint \( \PageIndex{ 4 } \)

Graph the ellipse given by the equation \(49x^2+16y^2=784\). Rewrite the equation in standard form. Then identify and label the center, vertices, co-vertices, and foci.

Graphing Ellipses Not Centered at the Origin

When an ellipse is not centered at the origin, we can still use the standard form to find the key features of the graph. When the ellipse is centered at some point, \((h, k)\), we use the standard form \(\frac{(x-h)^2}{v_x^2}+\frac{(y-k)^2}{v_y^2}=1 \). From this standard equation, we can easily determine the center, vertices, co-vertices, foci, and positions of the major and minor axes.

Exercise \(\PageIndex{5}\): Graphing an Ellipse Centered at \((h, k)\)

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

Solution

From the equation, \(v_x^2=4\) and \(v_y^2=9\). Moreover, \( h = -2 \) and \( k = 5 \). It follows that:

• the major axis is vertical (\(v_y^2 > v_x^2\))
• the center of the ellipse is \((-2,5)\)
• the coordinates of the vertices are \((-2, 5 \pm v_y)=(-2, 5 \pm \sqrt{9})=(-2, 5 \pm 3)\). That is, the vertices are at \( \left( -2,8 \right) \) and \( \left( -2,2 \right) \).
• the coordinates of the co-vertices are \((-2 \pm v_x, 5)=(-2 \pm \sqrt{4}, 5)=(-2 \pm 2, 5)\). Hence, they are at \( \left( -4,5 \right) \) and \( \left( 0,5 \right) \).
• the coordinates of the foci are \((-2, 5 \pm c)\), where \(c^2 = \left| v_x^2 - v_y^2 \right| \). Solving for \(c\), we have:\[ \begin{array}{rrcl} & c^2 & = & \left| 4 - 9 \right| \\[6pt] & & = & \left| -5 \right| \\[6pt] & & = & 5 \\[6pt] \implies & c & = & \sqrt{5} \\[6pt] \end{array} \nonumber \]

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.

Figure \(\PageIndex{7}\)
 

Checkpoint \( \PageIndex{ 5 } \)

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

Ellipses in General Form

We are sometimes given the equation of a conic in general form (not factored). These are equations of conics written in the form\[ a x^2 + b y^2 + c x +d y + e = 0, \nonumber \]where \( a \) and \( b \) are not both zero.

Definition: General Form of the Equation of a Conic Section

The general form for the equation of a conic section is \[ a x^2 + b y^2 + c x +d y + e = 0, \nonumber \]where \( a \) and \( b \) are not both zero.

To get equations like these into the celebrated standard form, we use our old friend - Completing the Square.

Example \(\PageIndex{6}\): Graphing an Ellipse Centered at \((h, k)\) by First Writing It in Standard Form

Graph the ellipse given by the equation \(4 x^2+9 y^2-40 x+36 y+100=0\). 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.\[ \left(4 x^2-40 x\right)+\left(9 y^2+36 y\right)=-100 \nonumber \]Factor out the coefficients of the squared terms.\[ 4\left(x^2-10 x\right)+9\left(y^2+4 y\right)=-100 \nonumber \]Complete the square twice. Remember to balance the equation by adding the same constants to each side.\[ 4\left(x^2-10 x+25\right)+9\left(y^2+4 y+4\right)=-100+100+36 \nonumber \]Rewrite as perfect squares.\[ 4(x-5)^2+9(y+2)^2=36 \nonumber \]Divide both sides by the constant term to place the equation in standard form.\[ \dfrac{(x-5)^2}{9}+\dfrac{(y+2)^2}{4}=1 \nonumber \]Now that the equation is in standard form, we can determine the position of the major axis. From the equation, \(v_x^2=9\) and \(v_y^2=4\). Moreover, \( h = 5 \) and \( k = -2 \). It follows that:

• the major axis is horizontal (\(v_x^2 > v_y^2\))
• the center of the ellipse is \((5,-2)\)
• the coordinates of the vertices are \((5 \pm v_x, -2)=(5 \pm \sqrt{9}, -2)=(5 \pm 3, -2)\). That is, \( \left( 2,-2 \right) \) and \( \left( 8,-2 \right) \)
• the coordinates of the co-vertices are \((5, -2 \pm v_y)=(5, -2 \pm \sqrt{4})=(5, -2 \pm 2)\). That is, \( \left( 5,-4 \right) \) and \( \left( 5,0 \right) \)
• the coordinates of the foci are \((5 \pm c,-2)\), where \(c^2 = \left| v_x^2 - v_y^2 \right| \). Solving for \(c\), we have:\[ \begin{array}{rrcl} & c^2 & = & \left| 9 - 4 \right| \\[6pt] & & = & \left| 5 \right| \\[6pt] & & = & 5 \\[6pt] \implies & c & = & \sqrt{5} \\[6pt] \end{array} \nonumber \]

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 in Figure \(\PageIndex{8}\).

Figure \(\PageIndex{8}\)
 

Checkpoint \( \PageIndex{ 6 } \)

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

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. See Figure \(\PageIndex{9}\). 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.

Figure \(\PageIndex{9}\)

Sound waves are reflected between foci in an elliptical room, called a whispering chamber.
 

Example \(\PageIndex{7}\): 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 as shown in Figure \(\PageIndex{10}\).

1. 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 \( (0,0) \).
2. 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.

Figure \(\PageIndex{10}\)
 

Solution

1. We are assuming a horizontal ellipse with center \((0,0)\), so we need to find an equation of the form \(\frac{x^2}{v_x^2}+\frac{y^2}{v_y^2}=1\), where \(v_x^2 > v_y^2\). We know that the length of the major axis, \(2 v_x\), is longer than the length of the minor axis, \(2 v_y\). 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 \(v_x\), we have \(2 v_x=96\), so \(v_x=48\), and \(v_x^2=2304\). Additionally, solving for \(v_y\), we have \(2 v_y=46\), so \(v_y=23\), and \(v_y^2=529\).

   Therefore, the equation of the ellipse is\[\dfrac{x^2}{2304}+\dfrac{y^2}{529}=1. \nonumber \]

2. To find the distance between the senators, we must find the distance between the foci, \(( \pm c, 0)\), where \(c^2= \left| a^2-b^2 \right|\). Solving for \(c\), we have:\[ \begin{array}{rrcl} & c^2 & = & \left| 2304 - 529 \right| \\[6pt] & & = & \left| 1775 \right| \\[6pt] & & = & 1775 \\[6pt] \implies & c & = & \sqrt{1775} \\[6pt] & & \approx & 42 \\[6pt] \end{array} \nonumber \]

Checkpoint \( \PageIndex{ 7 } \)

Suppose a whispering chamber is 480 feet long and 320 feet wide.

1. 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 \( (0,0) \).
2. 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