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.

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.

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 $\left(\left(\text{−}c,0\right),\left(c,0\right)\right)$

are on the x-axis and the center is the origin.

The definition states the sum of the distance from the foci to a point $\left(x,y\right)$

is constant. So $d_1+d_2$

is a constant that we will call $2a$

so, $d_1+d_2=2a.$

We will use the distance formula to lead us to an algebraic formula for an ellipse.

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

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.

We summarize the steps for reference.

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

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

We recognize this as the equation of an ellipse since both the x and y terms are squared and have different coefficients.	$x^2+4y^2=16$
To get the equation in standard form, divide both sides by 16 so that the equation is equal to 1.	$\frac{x^2}{16}+\frac{4y^2}{16}=\frac{16}{16}$
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 $\left(0,0\right).$
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 $\left(4,0\right),\left(\mathrm{-4},0\right).$ The endpoints of the minor axis are $\left(0,2\right),\left(0,\mathrm{-2}\right).$
Sketch the parabola.

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.

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 $\left(h,k\right).$

The equation is $\frac{{\left(x-h\right)}^2}{a^2}+\frac{{\left(y-k\right)}^2}{b^2}=1$

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

The equation is in standard form, $\frac{{\left(x-h\right)}^2}{a^2}+\frac{{\left(y-k\right)}^2}{b^2}=1.$	$\frac{{\left(x-3\right)}^2}{9}+\frac{{\left(y-1\right)}^2}{4}=1$
The ellipse is centered at $\left(h,k\right).$	The center is $\left(3,1\right).$
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 $\frac{{\left(x-3\right)}^2}{9}+\frac{{\left(y-1\right)}^2}{4}=1,$

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.

Notice in the graph above that we could have graphed $\frac{{\left(x-3\right)}^2}{9}+\frac{{\left(y-1\right)}^2}{4}=1$

by translations. We moved the original ellipse to the right 3 units and then up 1 unit.

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

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

by translation.

This ellipse will have the same size and shape as $\frac{x^2}{16}+\frac{y^2}{9}=1$

whose center is $\left(0,0\right).$

We graph this ellipse first.* * *

The center is $\left(0,0\right).$	Center $\left(0,0\right)$
Since $16>9,$ the major axis is horizontal.
$a^2=16,a=\pm 4$   $b^2=9,b=\pm 3$	The vertices are $\left(4,0\right),\left(\mathrm{-4},0\right).$ The endpoints of the minor axis are $\left(0,3\right),\left(0,\mathrm{-3}\right).$
Sketch the ellipse.
The original equation is in standard form, $\frac{{\left(x-h\right)}^2}{a^2}+\frac{{\left(y-k\right)}^2}{b^2}=1.$	$\frac{{\left(x-\left(\mathrm{-4}\right)\right)}^2}{16}+\frac{{\left(y-6\right)}^2}{9}=1$
The ellipse is centered at $\left(h,k\right).$	The center is $\left(\mathrm{-4},6\right).$
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 $\left(\mathrm{-4},6\right).$ The new ellipse is the ellipse whose equation is $\frac{{\left(x+4\right)}^2}{16}+\frac{{\left(y-6\right)}^2}{9}=1.$

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 $x^2+4y^2-4x+24y+24=0$

in standard form and graph.

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

	$x^2+4y^2-4x+24y+24=0$
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, $\frac{{\left(x-h\right)}^2}{a^2}+\frac{{\left(y-k\right)}^2}{b^2}=1$
The ellipse is centered at $\left(h,k\right).$	The center is $\left(2,\mathrm{-3}\right).$
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.

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 $\left(0,0\right):$

  The standard form of the equation of an ellipse with center

  $\left(0,0\right),$

  is

  ---

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

The x-intercepts are

and

The y-intercepts are

and

• How to an Ellipse with Center $\left(0,0\right)$
  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 $\left(h,k\right):$

  The standard form of the equation of an ellipse with center

  $\left(h,k\right),$

  is

  ---

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

When

the major axis is horizontal so the distance from the center to the vertex is a.

When

the major axis is vertical so the distance from the center to the vertex is b.

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