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Thread: Hyperbolas

  1. #1
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    Hyperbolas

    Could someone please tell me if the following HW problems are correct asap because I was not sure of them and there are no answers in the back of the book. Thanks!

    Give the following equations and points find the equation the hyperbola and graph:

    1) 9x^(2) -4y^(2)+54x +8y + 78= 0

    Answer: -(x+ 3)^(2) / 0.1 + (y-1) ^(2) / 0.25 = 1
    or
    (y-1) ^(2) / .25 - (x +3 ) ^(2) / 0.1 = 1

    2) Focus: (10, 0) Assymptotes: y= +/- 3/4x

    Answer: y^(2)/ 9 - x^(2) / 16 = 1
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  2. #2
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    Quote Originally Posted by googoogaga View Post
    Could someone please tell me if the following HW problems are correct asap because I was not sure of them and there are no answers in the back of the book. Thanks!

    Give the following equations and points find the equation the hyperbola and graph:

    1) 9x^(2) -4y^(2)+54x +8y + 78= 0

    Answer: -(x+ 3)^(2) / 0.1 + (y-1) ^(2) / 0.25 = 1
    or
    (y-1) ^(2) / .25 - (x +3 ) ^(2) / 0.1 = 1

    2) Focus: (10, 0) Assymptotes: y= +/- 3/4x

    Answer: y^(2)/ 9 - x^(2) / 16 = 1
    For the first one I get:
    $\displaystyle -\frac{(x + 3)^2}{\frac{1}{9}} + \frac{(y - 1)^2}{\frac{1}{4}} = 1$
    (Use fractions in the denominators. It will work better for you in the long run.)

    For the second one:
    $\displaystyle -\frac{x^2}{16} + \frac{y^2}{9} = 1$
    Graph it to check your work. (See below.)

    Everything looks good but the focus. (That little red dot out at (10, 0).) What can you think to do about that?

    -Dan
    Attached Thumbnails Attached Thumbnails Hyperbolas-hyperbola.jpg  
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  3. #3
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    Hello, googoogaga!

    Don't use decimals . . . and if you must, don't round them!


    Give the following equations and points, find the equation of the hyperbola and graph:

    1) 9x^2 - 4y^2 + 54x + 8y + 78 = 0

    Answer: . -(x+ 3) / 0.1 + (y-1) / 0.25 = 1
    . . . . . . . . . . . . . . . ?

    We have: .$\displaystyle 9(x^2 + 6x\qquad) - 4(y^2 - 2y\qquad) \;=\;-78$

    Complete the square: .$\displaystyle 9(x^2 + 6x \,{\color{blue}+\: 9})\: - 4(y^2-2y \:{\color{red}+\: 1})\;=\;-78 {\color{blue}\:+\: 81}\:{\color{red}\:-\: 4}$

    And we have: .$\displaystyle 9(x + 3)^2 - 4(y - 1)^2 \;=\;-1\quad\Rightarrow\quad 4(y-1)^2 - 9(x+3)^2\;=\;1$

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



    2) Focus: $\displaystyle (10,\,0)$ . Asymptotes: $\displaystyle y \:=\:\pm\frac{3}{4}x$

    Answer: .$\displaystyle \frac{y^2}{9} - \frac{x^2}{16} \;=\;1$. . . . no
    Since the asymptotes intersect at the origin, the center is at the origin.
    Since a focus is on the x-axis, we have a "horizontal" hyperbola.

    . . It has the form: .$\displaystyle \frac{x^2}{a^2} - \frac{y^2}{b^2} \:=\:1$

    Since the asymptotes are: .$\displaystyle y \:=\:\pm\frac{3}{4}x$, .we have: .$\displaystyle \frac{b}{a} \:=\:\frac{3}{4}\quad\Rightarrow\quad b \:=\:\frac{3}{4}a$

    The focal equation is: .$\displaystyle a^2 + b^2 \:=\:c^2$

    Since $\displaystyle c = 10$, we have: .$\displaystyle a^2 + \left(\frac{3}{4}a\right)^2 \:=\:10^2\quad\Rightarrow\quad \frac{25}{16}a^2\,=\,100\quad\Rightarrow\quad a \,=\,8$
    . . Then: .$\displaystyle b \,=\,6$

    The equation is: .$\displaystyle \frac{x^2}{64} - \frac{y^2}{36} \;=\;1$

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