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# Hyperbolas with any Center

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Hyperbolas Centered at (h, k)
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Your homework assignment is to graph the hyperbola $9(y + 2)^2 - 4(x -3)^2 = 36$ . What are the vertices of your graph?

### Guidance

Just like in the previous lessons, a hyperbola 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 $\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2}=1$ . We will address how the vertices, co-vertices, and foci change in the next example.

#### Example A

Graph $\frac{(x-2)^2}{16} - \frac{(y+1)^2}{9}=1$ . Then, find the vertices, foci, and asymptotes.

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

To graph, plot the center and then go out 4 units to the right and left and then up and down 3 units. Draw the box and asymptotes.

This is also how you can find the vertices. The vertices are $(2 \pm 4, -1)$ or $(6, -1)$ and $(-2, -1)$ .

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

$c^2 &= 16+9=25 \\c &= 5$

Therefore, the foci are $(2 \pm 5, -1)$ or $(7, -1)$ and $(-3, -1)$ .

To find the asymptotes, we have to do a little work to find the $y$ -intercepts. We know that the slope is $\pm \frac{b}{a}$ or $\pm \frac{3}{4}$ and they pass through the center. Let’s write each asymptote in point-slope form using the center and each slope.

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

Simplifying each equation, the asymptotes are $y=\frac{3}{4}x-\frac{5}{2}$ and $y=-\frac{3}{4}x+\frac{1}{2}$ .

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

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

#### Example B

Find the equation of the hyperbola with vertices $(-3, 2)$ and $(7, 2)$ and focus $(-5, 2)$ .

Solution: These two vertices create a horizontal transverse axis, making the hyperbola 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 $a$ , $|7 - 2|=5$ . The distance from the center to the given focus is $c$ , $|-5 -2|=7$ . Use $a$ and $c$ to solve for $b$ .

$7^2 &= 5^2+b^2 \\b^2 &= 24 \rightarrow b=2 \sqrt{6}$

Therefore, the equation is $\frac{(x-2)^2}{25} - \frac{(y-2)^2}{24}=1$ .

#### Example C

Graph $49(y-3)^2-25(x+4)^2=1225$ and find the foci.

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

$\frac{49(y-3)^2}{1225} - \frac{25(x+4)^2}{1225} &= \frac{1225}{1225} \\\frac{(y-3)^2}{25} - \frac{(x+4)^2}{49} &=1$

Now, we know that the hyperbola will be vertical because the $y$ -term is first. $a=5$ , $b=7$ and the center is $(-4, 3)$ .

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

$c^2 &= 49+25=74 \\c &= \sqrt{74}$

The foci are $\left(-4, 3 \pm \sqrt{74}\right)$ or $(-4, 11.6)$ and $(-4, -5.6)$ .

Intro Problem Revisit First we need to get the equation in standard form $\frac{(y-k)^2}{a^2} + \frac{(x-h)^2}{b^2} =1$ , so we divide by 36.

$9(y + 2)^2 - 4(x -3)^2 = 36\\\frac {9(y + 2)^2}{36} - \frac{4(x -3)^2}{36} = \frac{36}{36}\\\frac {(y + 2)^2}{4} - \frac{(x -3)^2}{9} = 1$ .

Because the y -term is first, we can now see that the vertices are $(h, k \pm a)$ $(3, -2 \pm 2)$ . That is, $(3, 0)$ and $(3, -4)$

### Guided Practice

1. Find the center, vertices, foci, and asymptotes of $\frac{(y-1)^2}{81} - \frac{(x+5)^2}{16}=1$ .

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

3. Find the equation of the hyperbola with vertices $(-6, -3)$ and $(-6, 5)$ and focus $(-6, 7)$ .

1. The center is $(-5, 1)$ , $a=\sqrt{81}=9$ and $b=\sqrt{16}=4$ , and the hyperbola is horizontal because the $y$ -term is first. The vertices are $\left(-5, 1 \pm 9\right)$ or $(-5, 10)$ and $(-5, -8)$ . Use $c^2=a^2+b^2$ to find $c$ .

$c^2 &= 81+16=97 \\c &= \sqrt{97}$

The foci are $\left(-5, 1+ \sqrt{97}\right)$ and $\left(-5, 1- \sqrt{97}\right)$ .

The asymptotes are $y-1= \pm \frac{9}{4}(x+5)$ or $y= \frac{9}{4}x + 12 \frac{1}{4}$ and $y= -\frac{9}{4}x - 10 \frac{1}{4}$ .

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

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

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

Find the foci.

$c^2 &= 25+4 = 29 \\c &= \sqrt{29}$

The foci are $\left(3, 1+ \sqrt{29}\right)$ and $\left(3, 1- \sqrt{29}\right)$ .

3. The vertices are $(-6, -3)$ and $(-6, 5)$ and the focus is $(-6, 7)$ . The transverse axis is going to be vertical because the $x$ -value does not change between these three points. The distance between the vertices is $|-3 -5|=8$ units, making $a=4$ . The midpoint between the vertices is the center.

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

The focus is $(-6, 7)$ and the distance between it and the center is 6 units, or $c$ . Find $b$ .

$36 &= b^2+16 \\20 &= b^2 \\b &= \sqrt{20}=2\sqrt{5}$

The equation of the hyperbola is $\frac{(y-1)^2}{16} - \frac{(x+6)^2}{20}=1$ .

### Vocabulary

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

### Practice

Find the center, vertices, foci, and asymptotes of each hyperbola below.

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

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

1. vertices: $(-2, -3)$ and $(8, -3)$ $b=7$
2. vertices: $(5, 6)$ and $(5, -12)$ focus: $(5, -15)$
3. asymptote: $y+3=\frac{4}{9}(x+1)$ horizontal transverse axis
4. foci: $(-11, -4)$ and $(1, -4)$ vertex: $(-8, -4)$
5. Extension Rewrite the equation of the hyperbola, $49x^2-4y^2+490x-16y+1013=0$ in standard form, by completing the square for both the $x$ and $y$ terms.