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BeccaB003

  • one year ago

Check my answer please? Thanks!!

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  1. BeccaB003
    • one year ago
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    Find the slopes of the asymptotes of a hyperbola with the equation y^2 = 36 + 4x^2 I got 1/4 and -1/4

  2. jim_thompson5910
    • one year ago
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    incorrect

  3. BeccaB003
    • one year ago
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    Then how do you solve? I can show you my steps if you'd like.

  4. jim_thompson5910
    • one year ago
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    sure go ahead and post them please

  5. jim_thompson5910
    • one year ago
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    I'll point out where you went wrong and then show how to do it

  6. BeccaB003
    • one year ago
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    okay, original equation: \[y^2 = 36 + 4x^2\] I put it in standard form: \[y^2-4x^2=36\] Then divide by 36 to get everything to equal 1 \[\frac{ y^2-4x^2=36 }{ 36 }\] \[\frac{ y^2 }{ 36 } - \frac{ x^2 }{ 9 }=1\] Slopes = \[\frac{ b }{ a }\] b = 3 a = 6 \[\frac{ 3 }{ 6 } = \frac{ 1 }{ 2 }\]

  7. BeccaB003
    • one year ago
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    hmm... doing it again I got 1/2.... I'm not sure what I did differently when I was typing out my steps because before I got 1/4...

  8. jim_thompson5910
    • one year ago
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    If you look at this page https://www.purplemath.com/modules/hyperbola.htm you'll notice what I'm attaching as an image

  9. jim_thompson5910
    • one year ago
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    a^2 = 36 so a = 6 b^2 = 9 so b = 3 slope = +-a/b = +-6/3 = +-2

  10. jim_thompson5910
    • one year ago
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    so you were very close when you got 1/2 and -1/2 you just had it flipped

  11. BeccaB003
    • one year ago
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    Wait, I thought the equation for slopes was b/a not a/b.... hmm... thanks for your help!! :D

  12. jim_thompson5910
    • one year ago
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    it depends on how the hyperbola is oriented, ie how it opens up (left/right vs top/bottom) that purplemath page describes it better with images

  13. jim_thompson5910
    • one year ago
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    Oh I see now

  14. jim_thompson5910
    • one year ago
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    for some reason, the 'a' and 'b' switch places the 'b' should be with the (y-k) term. I'm not sure why it went with the (x-h) term

  15. BeccaB003
    • one year ago
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    Oh! I see!! I always forget about how what equation you use is dependent on how the hyperbola is oriented.

  16. BeccaB003
    • one year ago
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    So, if the equations was going a different direction on the graph it would be b/a not a/b?

  17. jim_thompson5910
    • one year ago
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    yeah if the hyperbola is opening left/right, then you use b/a as the slope

  18. BeccaB003
    • one year ago
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    Thanks!

  19. jim_thompson5910
    • one year ago
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    you're welcome

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