Ellipses

Graph an Ellipse with Center at the Origin

The next conic section we will look at is an ellipse. We define an ellipse as all points in a plane where the sum of the distances from two fixed points is constant. Each of the given points is called a focus of the ellipse.

We can draw an ellipse by taking some fixed length of flexible string and attaching the ends to two thumbtacks. We use a pen to pull the string taut and rotate it around the two thumbtacks. The figure that results is an ellipse.

This figure shows a pen attached to two strings, the other ends of which are attached to two thumbtacks. The strings are pulled taut and the pen is rotated to draw an ellipse. The thumbtacks are labeled F subscript 1 and F subscript 2.

A line drawn through the foci intersect the ellipse in two points. Each point is called a vertex of the ellipse. The segment connecting the vertices is called the major axis. The midpoint of the segment is called the center of the ellipse. A segment perpendicular to the major axis that passes through the center and intersects the ellipse in two points is called the minor axis.

This figure shows two ellipses. In each, two points within the ellipse are labeled foci. A line drawn through the foci intersects the ellipse in two points. Each point is labeled a vertex. In the figure on the left, the segment connecting the vertices is called the major axis. A segment perpendicular to the major axis that passes through its midpoint and intersects the ellipse in two points is labeled minor axis. The minor axis is shorter than the minor axis. In the figure on the right, the segment through the foci, connecting the vertices is longer and is labeled major axis. Its midpoint is labeled center.

We mentioned earlier that our goal is to connect the geometry of a conic with algebra. Placing the ellipse on a rectangular coordinate system gives us that opportunity. In the figure, we placed the ellipse so the foci $((− c, 0), (c, 0))$ are on the x-axis and the center is the origin.

The figure on the left shows an ellipse with its center at the origin of the coordinate axes and its foci at points minus (c, 0) and (c, 0). A segment connects (negative c, 0) to a point (x, y) on the ellipse. The segment is labeled d subscript 1. Another segment, labeled d subscript 2 connects (c, 0) to (x, y). The figure on the right shows an ellipse with center at the origin, foci (negative c, 0) and (c, 0) and vertices (negative a, 0) and (a, 0). The point where the ellipse intersects the y axis is labeled (0, b). The segments connecting (0, 0) to (c, 0), (c, 0) to (0, b) and (0, b) to (0, 0) form a tight angled triangle with sides c, a and b respectively. The equation is a squared equals b squared plus c squared.

The definition states the sum of the distance from the foci to a point $(x, y)$ is constant. So $d_{1} + d_{2}$ is a constant that we will call $2 a$ so, $d_{1} + d_{2} = 2 a .$ We will use the distance formula to lead us to an algebraic formula for an ellipse.

\begin{matrix} \begin{matrix} Use the distance formula to find d_{1}, d_{2} . \end{matrix} & \begin{matrix} d_{1} & + & d_{2} & = & 2 a \\ \sqrt{{(x - (− c))}^{2} + {(y - 0)}^{2}} & + & \sqrt{{(x - c)}^{2} + {(y - 0)}^{2}} & = & 2 a \end{matrix} \\ \begin{matrix} After eliminating radicals and simplifying, \\ we get: \end{matrix} & \frac{x^{2}}{a^{2}} + \frac{y^{2}}{a^{2} - c^{2}} = 1 \\ \begin{matrix} To simplify the equation of the ellipse, we \\ let a^{2} - c^{2} = b^{2} . \end{matrix} \\ \begin{matrix} So, the equation of an ellipse centered at the \\ origin in standard form is: \end{matrix} & \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 \end{matrix}

To graph the ellipse, it will be helpful to know the intercepts. We will find the x-intercepts and y-intercepts using the formula.

\begin{matrix} \begin{matrix} y -intercepts \\ \begin{matrix} Let x = 0 . \end{matrix} & \begin{matrix} \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} & = & 1 \\ \frac{0^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} & = & 1 \\ \frac{y^{2}}{b^{2}} & = & 1 \\ y^{2} & = & b^{2} \\ y & = & \pm b \end{matrix} \end{matrix} & \begin{matrix} x -intercepts \\ \begin{matrix} Let y = 0 . \end{matrix} & \begin{matrix} \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} & = & 1 \\ \frac{x^{2}}{a^{2}} + \frac{0^{2}}{b^{2}} & = & 1 \\ \frac{x^{2}}{a^{2}} & = & 1 \\ x^{2} & = & a^{2} \\ x & = & \pm a \end{matrix} \end{matrix} \\ The y -intercepts are (0, b) and (0, − b) . & The x -intercepts are (a, 0) and (− a, 0) . \end{matrix}

Notice that when the major axis is horizontal, the value of a will be greater than the value of b and when the major axis is vertical, the value of b will be greater than the value of a. We will use this information to graph an ellipse that is centered at the origin.

Ellipse with Center $(0, 0)$
$\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$	$a > b$	$b > a$
Major axis	on the x- axis.	on the y-axis.
x-intercepts	$(− a, 0),$ $(a, 0)$
y-intercepts	$(0, − b),$ $(0, b)$

How to Graph an Ellipse with Center (0, 0)

Graph: $\frac{x^{2}}{4} + \frac{y^{2}}{9} = 1 .$

Solution

Step 1. Write the equation in standard form. It is in standard form x squared upon 6 plus y squared upon 9 equals 1. Step 2. Determine whether the major axis is horizontal or vertical. Since 9 is greater than 4 and 9 is in the y squared term, the major axis is vertical. Step 3. Find the endpoints of the major axis. The endpoints will be the y-intercepts. Since b squared is 9, b is plus or minus 3. The endpoints of the major axis are (0, 3) and (0, negative 3). Step 4. Find the endpoints of the minor axis. The endpoints will be the x-intercepts. Since a squared is 4, a is plus or minus 2. The endpoints of the minor axis are (2, 0) and (negative 2, 0). Step 5. Sketch the ellipse using the x and y intercepts. The graph shows an ellipse with center at (0, 0) and foci at (0, 3), (0, negative 3), (negative 2, 0), and (2, 0).

We summarize the steps for reference.

Sometimes our equation will first need to be put in standard form.

Graph $x^{2} + 4 y^{2} = 16 .$

Solution

We recognize this as the equation of an ellipse since both the x and y terms are squared and have different coefficients.	$x^{2} + 4 y^{2} = 16$
To get the equation in standard form, divide both sides by 16 so that the equation is equal to 1.	$\frac{x^{2}}{16} + \frac{4 y^{2}}{16} = \frac{16}{16}$
Simplify.	$\frac{x^{2}}{16} + \frac{y^{2}}{4} = 1$
The equation is in standard form. The ellipse is centered at the origin.	The center is $(0, 0) .$
Since $16 > 4$ and 16 is in the $x^{2}$ term, the major axis is horizontal.
$a^{2} = 16, a = \pm 4$ $b^{2} = 4, b = \pm 2$	The vertices are $(4, 0), (−4, 0) .$ The endpoints of the minor axis are $(0, 2), (0, −2) .$
Sketch the ellipse.

Find the Equation of an Ellipse with Center at the Origin

If we are given the graph of an ellipse, we can find the equation of the ellipse.

Find the equation of the ellipse shown.

This graph shows an ellipse with x intercepts (negative 4, 0) and (4, 0) and y intercepts (0, 3) and (0, negative 3).

Solution

Step-by-step derivation of an ellipse's standard equation by substituting its major and minor axis values.
We recognize this as an ellipse that is centered at the origin.	$\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$
Since the major axis is horizontal and the distance from the center to the vertex is 4, we know $a = 4$ and so $a^{2} = 16$ .	$\frac{x^{2}}{16} + \frac{y^{2}}{b^{2}} = 1$
The minor axis is vertical and the distance from the center to the ellipse is 3, we know $b = 3$ and so $b^{2} = 9$ .	$\frac{x^{2}}{16} + \frac{y^{2}}{9} = 1$

Graph an Ellipse with Center Not at the Origin

The ellipses we have looked at so far have all been centered at the origin. We will now look at ellipses whose center is $(h, k) .$

The equation is $\frac{{(x - h)}^{2}}{a^{2}} + \frac{{(y - k)}^{2}}{b^{2}} = 1$ and when $a > b,$ the major axis is horizontal so the distance from the center to the vertex is a. When $b > a,$ the major axis is vertical so the distance from the center to the vertex is b.

Graph: $\frac{{(x - 3)}^{2}}{9} + \frac{{(y - 1)}^{2}}{4} = 1 .$

Solution

The equation is in standard form, $\frac{{(x - h)}^{2}}{a^{2}} + \frac{{(y - k)}^{2}}{b^{2}} = 1 .$	$\frac{{(x - 3)}^{2}}{9} + \frac{{(y - 1)}^{2}}{4} = 1$
The ellipse is centered at $(h, k) .$	The center is $(3, 1) .$
Since $9 > 4$ and 9 is in the $x^{2}$ term, the major axis is horizontal.
$a^{2} = 9, a = \pm 3$ $b^{2} = 4, b = \pm 2$	The distance from the center to the vertices is 3. The distance from the center to the endpoints of the minor axis is 2.
Sketch the ellipse.

If we look at the equations of $\frac{x^{2}}{9} + \frac{y^{2}}{4} = 1$ and $\frac{{(x - 3)}^{2}}{9} + \frac{{(y - 1)}^{2}}{4} = 1,$ we see that they are both ellipses with $a = 3$ and $b = 2 .$ So they will have the same size and shape. They are different in that they do not have the same center.

The equation in the first figure is x squared upon 9 plus y squared upon 4 equals 1. Here, a is 3 and b is 2. The ellipse is graphed with center at (0, 0). The equation on the right is open parentheses x minus 3 close parentheses squared upon 9 plus open parentheses y minus 1 close parentheses squared upon 4 equals 1. Here, too, a is 3 and b is 2, but the center is (3, 1). The ellipse is shown on the same graph along with the first ellipse. The center is shown to have moved 3 units right and 1 unit up.

Notice in the graph above that we could have graphed $\frac{{(x - 3)}^{2}}{9} + \frac{{(y - 1)}^{2}}{4} = 1$ by translations. We moved the original ellipse to the right 3 units and then up 1 unit.

This graph shows an ellipse translated from center (0, 0) to center (3, 1). The center has moved 3 units right and 1 unit up. The original ellipse has vertices at (negative 3, 0) and (3, 0) and endpoint of minor axis at (negative 2, 0) and (2, 0). The translated ellipse has vertices at (0, 1) and (6, 1) and endpoints of minor axis at (3, negative 1) and (3, 3).

In the next example we will use the translation method to graph the ellipse.

Graph $\frac{{(x + 4)}^{2}}{16} + \frac{{(y - 6)}^{2}}{9} = 1$ by translation.

Solution

This ellipse will have the same size and shape as

\frac{x^{2}}{16} + \frac{y^{2}}{9} = 1

whose center is

(0, 0) .

We graph this ellipse first.

The center is $(0, 0) .$	Center $(0, 0)$
Since $16 > 9,$ the major axis is horizontal.
$a^{2} = 16, a = \pm 4$ $b^{2} = 9, b = \pm 3$	The vertices are $(4, 0), (−4, 0) .$ The endpoints of the minor axis are $(0, 3), (0, −3) .$
Sketch the ellipse.
The original equation is in standard form, $\frac{{(x - h)}^{2}}{a^{2}} + \frac{{(y - k)}^{2}}{b^{2}} = 1 .$	$\frac{{(x - (−4))}^{2}}{16} + \frac{{(y - 6)}^{2}}{9} = 1$
The ellipse is centered at $(h, k) .$	The center is $(−4, 6) .$
We translate the graph of $\frac{x^{2}}{16} + \frac{y^{2}}{9} = 1$ four units to the left and then up 6 units. Verify that the center is $(−4, 6) .$ The new ellipse is the ellipse whose equation is $\frac{{(x + 4)}^{2}}{16} + \frac{{(y - 6)}^{2}}{9} = 1 .$

When an equation has both an $x^{2}$ and a $y^{2}$ with different coefficients, we verify that it is an ellipsis by putting it in standard form. We will then be able to graph the equation.

Write the equation $x^{2} + 4 y^{2} - 4 x + 24 y + 24 = 0$ in standard form and graph.

Solution

We put the equation in standard form by completing the squares in both x and y.

	$x^{2} + 4 y^{2} - 4 x + 24 y + 24 = 0$
Rewrite grouping the x terms and y terms.
Make the coefficients of $x^{2}$ and $y^{2}$ equal 1.
Complete the squares.
Write as binomial squares.
Divide both sides by 16 to get 1 on the right.
Simplify.
The equation is in standard form, $\frac{{(x - h)}^{2}}{a^{2}} + \frac{{(y - k)}^{2}}{b^{2}} = 1$
The ellipse is centered at $(h, k) .$	The center is $(2, −3) .$
Since $16 > 4$ and 16 is in the $x^{2}$ term, the major axis is horizontal. $a^{2} = 16, a = \pm 4$ $b^{2} = 4, b = \pm 2$	The distance from the center to the vertices is 4. The distance from the center to the endpoints of the minor axis is 2.
Sketch the ellipse.

Solve Application with Ellipses

The orbits of the planets around the sun follow elliptical paths.

Pluto (a dwarf planet) moves in an elliptical orbit around the Sun. The closest Pluto gets to the Sun is approximately 30 astronomical units (AU) and the furthest is approximately 50 AU. The Sun is one of the foci of the elliptical orbit. Letting the ellipse center at the origin and labeling the axes in AU, the orbit will look like the figure below. Use the graph to write an equation for the elliptical orbit of Pluto.

This graph shows an ellipse with center (0, 0) and vertices (negative 40, 0) and (40, 0). The sun is shown at point (10, 0). This is 30 units from the right vertex and 50 units from the left vertex.

Solution

Steps to determine the standard form equation of an ellipse by identifying parameters like axes and foci, and performing relevant calculations.
We recognize this as an ellipse that is centered at the origin.	$\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$
Since the major axis is horizontal and the distance from the center to the vertex is 40, we know $a = 40$ and so $a^{2} = 1600$ .	$\frac{x^{2}}{1600} + \frac{y^{2}}{b^{2}} = 1$
The minor axis is vertical but the end points aren’t given. To find $b$ we will use the location of the Sun. Since the Sun is a focus of the ellipse at the point $(10, 0)$ , we know $c = 10$ . Use this to solve for $b^{2}$ .	$\begin{matrix} b^{2} = a^{2} - c^{2} \\ b^{2} = 40^{2} - 10^{2} \\ b^{2} = 1600 - 100 \\ b^{2} = 1500 \end{matrix}$
Substitute $a^{2}$ and $b^{2}$ into the standard form of the ellipse.	$\frac{x^{2}}{1600} + \frac{y^{2}}{1500} = 1$

Key Concepts

Ellipse: An ellipse is all points in a plane where the sum of the distances from two fixed points is constant. Each of the fixed points is called a focus of the ellipse.

If we draw a line through the foci intersects the ellipse in two points—each is called a vertex of the ellipse.
The segment connecting the vertices is called the major axis.
The midpoint of the segment is called the center of the ellipse.
A segment perpendicular to the major axis that passes through the center and intersects the ellipse in two points is called the minor axis.
Standard Form of the Equation an Ellipse with Center $(0, 0) :$ The standard form of the equation of an ellipse with center $(0, 0),$ is

$\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$

The x-intercepts are $(a, 0)$ and $(− a, 0) .$
The y-intercepts are $(0, b)$ and $(0, − b) .$
How to an Ellipse with Center (0,0)
1. Write the equation in standard form.
2. Determine whether the major axis is horizontal or vertical.
3. Find the endpoints of the major axis.
4. Find the endpoints of the minor axis
5. Sketch the ellipse.
Standard Form of the Equation an Ellipse with Center $(h, k) :$ The standard form of the equation of an ellipse with center $(h, k),$ is

$\frac{{(x - h)}^{2}}{a^{2}} + \frac{{(y - k)}^{2}}{b^{2}} = 1$

When $a > b,$ the major axis is horizontal so the distance from the center to the vertex is a.
When $b > a,$ the major axis is vertical so the distance from the center to the vertex is b.

Practice Makes Perfect

Graph an Ellipse with Center at the Origin

In the following exercises, graph each ellipse.

$\frac{x^{2}}{4} + \frac{y^{2}}{25} = 1$

Solution

This graph shows an ellipse with center (0, 0), vertices (0, 5) and (0, negative 5) and endpoints of minor axis (2, 0) and (negative 2, 0).

$\frac{x^{2}}{9} + \frac{y^{2}}{25} = 1$

$\frac{x^{2}}{25} + \frac{y^{2}}{36} = 1$

Solution

This graph shows an ellipse with center (0, 0), vertices (0, 6) and (0, negative 6) and endpoints of minor axis (5, 0) and (negative 5, 0).

$\frac{x^{2}}{16} + \frac{y^{2}}{36} = 1$

$\frac{x^{2}}{36} + \frac{y^{2}}{16} = 1$

Solution

This graph shows an ellipse with center (0, 0), vertices (6, 0) and (negative 6, 0) and endpoints of minor axis (0, 4) and (0, negative 4).

$\frac{x^{2}}{25} + \frac{y^{2}}{9} = 1$

$x^{2} + \frac{y^{2}}{4} = 1$

Solution

This graph shows an ellipse with center (0, 0), vertices (0, 2) and (0, negative 2) and endpoints of minor axis (1, 0) and (negative 1, 0).

$\frac{x^{2}}{9} + y^{2} = 1$

$4 x^{2} + 25 y^{2} = 100$

Solution

This graph shows an ellipse with center (0, 0), vertices (5, 0) and (negative 5, 0) and endpoints of minor axis (0, 2) and (0, negative 2).

$16 x^{2} + 9 y^{2} = 144$

$16 x^{2} + 36 y^{2} = 576$

Solution

This graph shows an ellipse with center (0, 0), vertices (6, 0) and (negative 6, 0) and endpoints of minor axis (0, 4) and (0, negative 4).

$9 x^{2} + 25 y^{2} = 225$

Find the Equation of an Ellipse with Center at the Origin

In the following exercises, find the equation of the ellipse shown in the graph.

This graph shows an ellipse with center (0, 0), vertices (0, 5) and (0, negative 5) and endpoints of minor axis (negative 3, 0) and (3, 0).

Solution

$\frac{x^{2}}{9} + \frac{y^{2}}{25} = 1$

This graph shows an ellipse with center (0, 0), vertices (5, 0) and (negative 5, 0) and endpoints of minor axis (0, 2) and (0, negative 2).

This graph shows an ellipse with center (0, 0), vertices (0, 4) and (0, negative 4) and endpoints of minor axis (negative 3, 0) and (3, 0).

Solution

$\frac{x^{2}}{9} + \frac{y^{2}}{16} = 1$

This graph shows an ellipse with center (0, 0), vertices (0, 6) and (0, negative 6) and endpoints of minor axis (negative 4, 0) and (4, 0).

Graph an Ellipse with Center Not at the Origin

In the following exercises, graph each ellipse.

$\frac{{(x + 1)}^{2}}{4} + \frac{{(y + 6)}^{2}}{25} = 1$

Solution

This graph shows an ellipse with center (negative 1, negative 6, vertices (negative 1, negative 1) and (negative 1, negative 11) and endpoints of minor axis (negative 3, negative 6) and (1, negative 6).

$\frac{{(x - 3)}^{2}}{25} + \frac{{(y + 2)}^{2}}{9} = 1$

$\frac{{(x + 4)}^{2}}{4} + \frac{{(y - 2)}^{2}}{9} = 1$

Solution

This graph shows an ellipse with center (negative 4, 2, vertices (negative 4, 5) and (negative 4, negative 1) and endpoints of minor axis (3, 1) and (negative 6, 2) and (negative 2, 2).

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

In the following exercises, graph each equation by translation.

$\frac{{(x - 3)}^{2}}{4} + \frac{{(y - 7)}^{2}}{25} = 1$

Solution

This graph shows an ellipse with center (3, 7), vertices (3, 2) and (3, 12), and endpoints of minor axis (1, 7) and (5, 7).

$\frac{{(x + 6)}^{2}}{16} + \frac{{(y + 5)}^{2}}{4} = 1$

$\frac{{(x - 5)}^{2}}{9} + \frac{{(y + 4)}^{2}}{25} = 1$

Solution

This graph shows an ellipse with center (5, negative 4), vertices (5, 1) and (5, negative 9) and endpoints of minor axis (2, negative 4) and (8, negative 4).

$\frac{{(x + 5)}^{2}}{36} + \frac{{(y - 3)}^{2}}{16} = 1$

In the following exercises, ⓐ write the equation in standard form and ⓑ graph.

$25 x^{2} + 9 y^{2} - 100 x - 54 y - 44 = 0$

Solution

ⓐ $\frac{{(x - 2)}^{2}}{9} + \frac{{(y - 3)}^{2}}{25} = 1$
ⓑ
This graph shows an ellipse with center (2, 3), vertices (2, negative 2) and (2, 8) and endpoints of minor axis (negative 1, 3) and (5, 3).

$4 x^{2} + 25 y^{2} + 8 x + 100 y + 4 = 0$

$4 x^{2} + 25 y^{2} - 24 x - 64 = 0$

Solution

ⓐ $\frac{y^{2}}{4} + \frac{{(x - 3)}^{2}}{25} = 1$
ⓑ
This graph shows an ellipse with center (3, 0), vertices (negative 2, 0) and (8, 0) and endpoints of minor axis (3, 2) and (3, negative 2).

$9 x^{2} + 4 y^{2} + 56 y + 160 = 0$

In the following exercises, graph the equation.

$x = −2 {(y - 1)}^{2} + 2$

Solution

This graph shows a parabola with vertex (2, 1) and y intercepts (0, 0) and (2, 0).

$x^{2} + y^{2} = 49$

${(x + 5)}^{2} + {(y + 2)}^{2} = 4$

Solution

This graph shows a circle with center (negative 5, negative 2) and a radius of 2 units.

$y = − x^{2} + 8 x - 15$

$\frac{{(x + 3)}^{2}}{16} + \frac{{(y + 1)}^{2}}{4} = 1$

Solution

This graph shows an ellipse with center (negative 3, negative 1), vertices (1, negative 1) and (negative 7, negative 1) and endpoints of minor axis (negative 3, 1) and (negative 3, negative 3).

${(x - 2)}^{2} + {(y - 3)}^{2} = 9$

$\frac{x^{2}}{25} + \frac{y^{2}}{36} = 1$

Solution

This graph shows an ellipse with center (0, 0), vertices (0, 6) and (0, negative 6) and endpoints of minor axis (negative 5, 0) and (5, 0).

$x = 4 {(y + 1)}^{2} - 4$

$x^{2} + y^{2} = 64$

Solution

This graph shows circle with center (0, 0) and with radius 8 units.

$\frac{x^{2}}{9} + \frac{y^{2}}{25} = 1$

$y = 6 x^{2} + 2 x - 1$

Solution

This graph shows upward opening parabola. Its vertex has an x value of slightly less than 0 and a y value of slightly less than minus 1. A point on it is approximately at (negative 1, 3).

$\frac{{(x - 2)}^{2}}{9} + \frac{{(y + 3)}^{2}}{25} = 1$

Solve Application with Ellipses

A planet moves in an elliptical orbit around its sun. The closest the planet gets to the sun is approximately 10 AU and the furthest is approximately 30 AU. The sun is one of the foci of the elliptical orbit. Letting the ellipse center at the origin and labeling the axes in AU, the orbit will look like the figure below. Use the graph to write an equation for the elliptical orbit of the planet.

This graph shows an ellipse with center (0, 0), vertices (negative 20, 0) and (20, 0). The sun is shown at point (10, 0), which is 30 units from the left vertex and 10 units from the right vertex.

Solution

$\frac{x^{2}}{400} + \frac{y^{2}}{300} = 1$

A planet moves in an elliptical orbit around its sun. The closest the planet gets to the sun is approximately 10 AU and the furthest is approximately 70 AU. The sun is one of the foci of the elliptical orbit. Letting the ellipse center at the origin and labeling the axes in AU, the orbit will look like the figure below. Use the graph to write an equation for the elliptical orbit of the planet.

This graph shows an ellipse with center (0, 0), vertices (negative 40, 0) and (40, 0). The sun is shown at point (30, 0), which is 70 units from the left vertex and 10 units from the right vertex.

A comet moves in an elliptical orbit around a sun. The closest the comet gets to the sun is approximately 15 AU and the furthest is approximately 85 AU. The sun is one of the foci of the elliptical orbit. Letting the ellipse center at the origin and labeling the axes in AU, the orbit will look like the figure below. Use the graph to write an equation for the elliptical orbit of the comet.

This graph shows an ellipse with center (0, 0), vertices (negative 50, 0) and (50, 0). The sun is shown at point (35, 0), which is 85 units from the left vertex and 15 units from the right vertex.

Solution

$\frac{x^{2}}{2500} + \frac{y^{2}}{1275} = 1$

A comet moves in an elliptical orbit around a sun. The closest the comet gets to the sun is approximately 15 AU and the furthest is approximately 95 AU. The sun is one of the foci of the elliptical orbit. Letting the ellipse center at the origin and labeling the axes in AU, the orbit will look like the figure below. Use the graph to write an equation for the elliptical orbit of the comet.

This graph shows an ellipse with center (0, 0), vertices (negative 55, 0) and (55, 0). The sun is shown at point (40, 0), which is 95 units from the left vertex and 15 units from the right vertex.

Writing Exercises

In your own words, define an ellipse and write the equation of an ellipse centered at the origin in standard form. Draw a sketch of the ellipse labeling the center, vertices and major and minor axes.

Solution

Answers will vary.

Explain in your own words how to get the axes from the equation in standard form.

Compare and contrast the graphs of the equations $\frac{x^{2}}{4} + \frac{y^{2}}{9} = 1$ and $\frac{x^{2}}{9} + \frac{y^{2}}{4} = 1 .$

Solution

Answers will vary.

Explain in your own words, the difference between a vertex and a focus of the ellipse.

Self Check

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

This table has 4 columns 4 rows and a header row. The header row labels each column I can, confidently, with some help and no, I don’t get it. The first columns has the following statements: graph an ellipse with center at the origin, find the equation of an ellipse with center at the origin, graph an ellipse with center not at the origin, solve applications with ellipses. The remaining columns are blank.

ⓑ What does this checklist tell you about your mastery of this section? What steps will you take to improve?

ellipse

An ellipse is all points in a plane where the sum of the distances from two fixed points is constant.