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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) .

Answers

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.

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