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anonymous

  • one year ago

Find an equation in standard form for the hyperbola with vertices at (0, ±10) and asymptotes at y = ±5/6* x.

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  1. Plasmataco
    • one year ago
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    ... I'll try but no garentees

  2. anonymous
    • one year ago
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    No problem! Atleast if you can try or get someone to help you too

  3. Plasmataco
    • one year ago
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    It's center is at the origin so...

  4. Plasmataco
    • one year ago
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    Ok... U listening?

  5. Plasmataco
    • one year ago
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    What u got to do is put the x in the coordinates of the vertices in for the asymptote equation.

  6. Plasmataco
    • one year ago
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    So plug 10 in to get y=50/6

  7. anonymous
    • one year ago
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    is this vertical or horiztonal though

  8. Plasmataco
    • one year ago
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    .

  9. Plasmataco
    • one year ago
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    Hang on.

  10. Plasmataco
    • one year ago
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    Because the center is (0,0), it makes thing a lot easier

  11. anonymous
    • one year ago
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    Oh okay

  12. Plasmataco
    • one year ago
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    The equation is ((X-g)/a)-((y-f)/b)=1

  13. anonymous
    • one year ago
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    okay yeah I have that

  14. Plasmataco
    • one year ago
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    (G,f) is the center of the hyperbola in which case (0,0) so we can cancel those out to get (X squared)/a-(y squared)/b

  15. Plasmataco
    • one year ago
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    My first equation is wrong. I forgot the power 2 after(x-g) and (y-f) sry 😓

  16. anonymous
    • one year ago
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    okay yeah I was gonna correct that

  17. Plasmataco
    • one year ago
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    A is the x part of the vertice so 10. B is what we solved earlier with the asymptote.

  18. anonymous
    • one year ago
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    wait what is b?

  19. Plasmataco
    • one year ago
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    So we get (x squared)/10-(y squared times 6)/50=1

  20. Plasmataco
    • one year ago
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    The 6th comment

  21. Plasmataco
    • one year ago
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    No wait the answer is (x/10) squared-(6y/50) squared=1

  22. Plasmataco
    • one year ago
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    Forgot another pair of squares at the denominator

  23. Plasmataco
    • one year ago
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    Here's the link I used

  24. Plasmataco
    • one year ago
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    http://www.endmemo.com/geometry/hyperbola.php

  25. Plasmataco
    • one year ago
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    @PHUNISH ?

  26. Plasmataco
    • one year ago
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    Ok well. I gotta go sry

  27. Plasmataco
    • one year ago
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    I hoped I helped

  28. anonymous
    • one year ago
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    Yeah sorry I been trying to figure this out with my notes but you helped so much more thank you so much

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