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# Ellipses Not Centered at the Origin

## Write and graph equations of ellipses with centers not at (0, 0)

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Practice Ellipses Not Centered at the Origin
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Ellipses Centered at (h, k)

Your homework assignment is to draw the ellipse $16(x-2)^2+4(y+3)^2=144$ . What is the vertex of your graph and where will the foci of the ellipse be located?

### Guidance

Just like in the previous lessons, an ellipse does not always have to be placed with its center at the origin. If the center is $(h, k)$ the entire ellipse will be shifted $h$ units to the left or right and $k$ units up or down. The equation becomes $\frac{\left(x-h\right)^2}{a^2}+ \frac{\left(y-k\right)^2}{b^2}=1$ . We will address how the vertices, co-vertices, and foci change in the next example.

#### Example A

Graph $\frac{\left(x-3\right)^2}{16}+ \frac{\left(y+1\right)^2}{4}=1$ . Then, find the vertices, co-vertices, and foci.

Solution: First, we know this is a horizontal ellipse because $16 > 4$ . Therefore, the center is $(3, -1)$ and $a = 4$ and $b = 2$ . Use this information to graph the ellipse.

To graph, plot the center and then go out 4 units to the right and left and then up and down two units. This is also how you can find the vertices and co-vertices. The vertices are $(3 \pm 4,-1)$ or $(7, -1)$ and $(-1, -1)$ . The co-vertices are $(3,-1 \pm 2)$ or $(3, 1)$ and $(3, -3)$ .

To find the foci, we need to find $c$ using $c^2=a^2-b^2$ .

$c^2&=16-4=12 \\c&=2\sqrt{3}$

Therefore, the foci are $\left(3 \pm 2\sqrt{3},-1\right)$ .

From this example, we can create formulas for finding the vertices, co-vertices, and foci of an ellipse with center $(h, k)$ . Also, when graphing an ellipse, not centered at the origin, make sure to plot the center.

Orientation Equation Vertices Co-Vertices Foci
Horizontal $\frac{\left(x-h\right)^2}{a^2}+ \frac{\left(y-k\right)^2}{b^2}=1$ $(h \pm a,k)$ $(h,k \pm b)$ $(h \pm c,k)$
Vertical $\frac{\left(x-h\right)^2}{b^2}+ \frac{\left(y-k\right)^2}{a^2}=1$ $(h,k \pm a)$ $(h \pm b,k)$ $(h,k \pm c)$

#### Example B

Find the equation of the ellipse with vertices $(-3, 2)$ and $(7, 2)$ and co-vertex $(2, -1)$ .

Solution: These two vertices create a horizontal major axis, making the ellipse horizontal. If you are unsure, plot the given information on a set of axes. To find the center, use the midpoint formula with the vertices.

$\left(\frac{-3+7}{2}, \frac{2+2}{2}\right)= \left(\frac{4}{2}, \frac{4}{2}\right)=(2,2)$

The distance from one of the vertices to the center is $a$ , $\left |7-2 \right \vert=5$ . The distance from the co-vertex to the center is $b$ , $\left |-1-2 \right \vert=3$ . Therefore, the equation is $\frac{\left(x-2\right)^2}{5^2}+ \frac{\left(y-2\right)^2}{3^2}=1$ or $\frac{\left(x-2\right)^2}{25}+ \frac{\left(y-2\right)^2}{9}=1$ .

#### Example C

Graph $49(x-5)^2+25(y+2)^2=1225$ and find the foci.

Solution: First we have to get this into standard form, like the equations above. To make the right side 1, we need to divide everything by 1225.

$\frac{49\left(x-5\right)^2}{1225}+ \frac{25\left(y+2\right)^2}{1225}&=\frac{1225}{1225} \\\frac{\left(x-5\right)^2}{25}+ \frac{\left(y+2\right)^2}{49}&=1$

Now, we know that the ellipse will be vertical because $25 < 49$ . $a = 7, b = 5$ and the center is $(5, -2)$ .

To find the foci, we first need to find $c$ by using $c^2=a^2-b^2$ .

$c^2&=49-25=24 \\c&=\sqrt{24}=2\sqrt{6}$

The foci are $\left(5,-2 \pm 2\sqrt{6}\right)$ or $(5, -6.9)$ and $(5, 2.9)$ .

Intro Problem Revisit We first need to get our equation in the form of $\frac{\left(x-h\right)^2}{a^2}+ \frac{\left(y-k\right)^2}{b^2}=1$ . So we divide both sides by 144.

$\frac{16(x-2)^2}{144} + \frac{4(y+3)^2}{144} = \frac{144}{144}\\\frac{(x-2)^2}{9} + \frac{(y + 3)^2}{36}$ .

Now we can see that $h = 2$ and $3 = -k$ or $k = -3$ . Therefore the origin is $(2, -3)$ .

Because $9 < 36$ , we know this is a vertical ellipse. To find the foci, use $c^2=a^2-b^2$ .

$c^2&=36-9=27 \\c&=\sqrt{27}=3\sqrt{3}$

The foci are therefore $\left(0,3\sqrt{3}\right)$ and $\left(0,-3\sqrt{3}\right)$ .

### Guided Practice

1. Find the center, vertices, co-vertices and foci of $\frac{\left(x+4\right)^2}{81}+ \frac{\left(y-7\right)^2}{16}=1$ .

2. Graph $25(x-3)^2+4(y-1)^2=100$ and find the foci.

3. Find the equation of the ellipse with co-vertices $(-3, -6)$ and $(5, -6)$ and focus $(1, -2)$ .

1. The center is $(-4, 7),a=\sqrt{81}=9$ and $b=\sqrt{16}=4$ , making the ellipse horizontal. The vertices are $(-4 \pm 9,7)$ or $(-13, 7)$ and $(5, 7)$ . The co-vertices are $(-4,7 \pm 4)$ or $(-4, 3)$ and $(-4, 11)$ . Use $c^2=a^2-b^2$ to find $c$ .

$c^2&=81-16=65\\c&=\sqrt{65}$

The foci are $\left(-4- \sqrt{65},7\right)$ and $\left(-4+ \sqrt{65},7\right)$ .

2. Change this equation to standard form in order to graph.

$\frac{25\left(x-3\right)^2}{100}+ \frac{4\left(y-1\right)^2}{100}&=\frac{100}{100} \\\frac{\left(x-3\right)^2}{4}+ \frac{\left(y-1\right)^2}{25}&=1$

center: $(3, 1), b = 2, a = 5$

Find the foci.

$c^2&=25-4=21 \\c&=\sqrt{21}$

The foci are $\left(3,1+ \sqrt{21}\right)$ and $\left(3,1- \sqrt{21}\right)$ .

3. The co-vertices $(-3, -6)$ and $(5, -6)$ are the endpoints of the minor axis. It is $\left |-3-5\right \vert=8$ units long, making $b = 4$ . The midpoint between the co-vertices is the center.

$\left(\frac{-3+5}{2},-6\right)=\left(\frac{2}{2},-6\right)=(1,-6)$

The focus is $(1, -1)$ and the distance between it and the center is 4 units, or $c$ . Find $a$ .

$16&=a^2-16 \\32&=a^2 \\a&=\sqrt{32}=4\sqrt{2}$

The equation of the ellipse is $\frac{\left(x-1\right)^2}{16}+ \frac{\left(y+6\right)^2}{32}=1$ .

### Vocabulary

Standard Form (of an Ellipse)
$\frac{\left(x-h\right)^2}{a^2}+ \frac{\left(y-k\right)^2}{b^2}=1$ or $\frac{\left(x-h\right)^2}{b^2}+ \frac{\left(y-k\right)^2}{a^2}=1$ where $(h, k)$ is the center.

### Practice

Find the center, vertices, co-vertices, and foci of each ellipse below.

1. $\frac{\left(x+5\right)^2}{25}+ \frac{\left(y+1\right)^2}{36}=1$
2. $(x+2)^2+16(y-6)^2=16$
3. $\frac{\left(x-2\right)^2}{9}+\frac{\left(y-3\right)^2}{49}=1$
4. $25x^2+64(y-6)^2=1600$
5. $(x-8)^2+ \frac{\left(y-4\right)^2}{9}=1$
6. $81(x+4)^2+4(y+5)^2=324$
7. Graph the ellipse in #1.
8. Graph the ellipse in #2.
9. Graph the ellipse in #4.
10. Graph the ellipse in #5.

Using the information below, find the equation of each ellipse.

1. vertices: $(-2, -3)$ and $(8, -3)$ co-vertex: $(3, -5)$
2. vertices: $(5, 6)$ and $(5, -12)$ focus: $(5, -7)$
3. co-vertices: $(0, 4)$ and $(14, 4)$ focus: $(7, 1)$
4. foci: $(-11, -4)$ and $(1, -4)$ vertex: $(-12, -4)$
5. Extension Rewrite the equation of the ellipse, $36x^2+25y^2-72x+200y-464=0$ in standard form, by completing the square for both the $x$ and $y$ terms.