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anonymous

  • one year ago

Find an equation in standard form for the hyperbola with vertices at (0, ±8) and asymptotes at y equals inverse of plus minus 1 divided by 2x.

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  1. anonymous
    • one year ago
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    @jim_thompson5910

  2. jim_thompson5910
    • one year ago
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    what is the center of this hyperbola?

  3. anonymous
    • one year ago
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    0,0

  4. jim_thompson5910
    • one year ago
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    yes

  5. jim_thompson5910
    • one year ago
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    so (h,k) = (0,0) h = 0 k = 0

  6. anonymous
    • one year ago
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    hey that makes a cool face 0,0

  7. anonymous
    • one year ago
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    sorry off topic

  8. jim_thompson5910
    • one year ago
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    lol yes it does how far do go from (0,0) to either vertex?

  9. anonymous
    • one year ago
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    8 units

  10. jim_thompson5910
    • one year ago
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    so a = 8 a = half of the length of the transverse axis the transverse axis connects the two vertices and goes through the center

  11. anonymous
    • one year ago
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    alright

  12. anonymous
    • one year ago
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    SO what is the formula that we need to plug these into?

  13. jim_thompson5910
    • one year ago
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    how is the hyperbola oriented? is it opening left/right? or up/down?

  14. anonymous
    • one year ago
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    up/down

  15. jim_thompson5910
    • one year ago
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    according to this http://www.purplemath.com/modules/hyperbola.htm it says that if the hyperbola opens up/down, then the basic form is \[\Large \frac{(y-k)^2}{a^2} + \frac{(x-h)^2}{b^2} = 1\]

  16. anonymous
    • one year ago
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    yup

  17. jim_thompson5910
    • one year ago
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    on that same page it says that the asymptotes have a slope of +-a/b focus on say the positive slope slope = 1/2 a/b = 1/2 8/b = 1/2 b = ??

  18. anonymous
    • one year ago
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    16

  19. jim_thompson5910
    • one year ago
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    \[\Large \frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1\] \[\Large \frac{(y-0)^2}{8^2} - \frac{(x-0)^2}{16^2} = 1\] \[\Large \frac{y^2}{64} - \frac{x^2}{256} = 1\]

  20. jim_thompson5910
    • one year ago
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    I had a + in there for some reason. It should have been -

  21. anonymous
    • one year ago
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    Ok thanks! I have 4 more questions of similar nature if you dont't mind :)

  22. anonymous
    • one year ago
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    I'll make new questions

  23. jim_thompson5910
    • one year ago
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    try the next one out and I'll take a look at the work you have so far on it

  24. anonymous
    • one year ago
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    Not the same type of question but the same chapter in my book

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