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anonymous

  • 5 years ago

Find the vertices and asymptotes of the hyperbola. 25y^2 - 16x^2 = 400

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  1. anonymous
    • 5 years ago
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    \[y^2/(400/25) - x^2/(400/16) =1\]

  2. anonymous
    • 5 years ago
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    \[y^2/(4)^2- x^2/(5)^2= 1\]

  3. anonymous
    • 5 years ago
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    shouldnt it be 25/400 and not the other way around

  4. anonymous
    • 5 years ago
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    no 25 is multiplied with y^2

  5. anonymous
    • 5 years ago
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    y=(+/-)b/a x are the asymptotes

  6. anonymous
    • 5 years ago
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    ok so the asymptote.. y=+-4/5x

  7. anonymous
    • 5 years ago
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    then does that mean the vertices are (0,+-4)

  8. anonymous
    • 5 years ago
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    Yes uxma but as you are taking 400 from the left hand side it should be the denominator in this case

  9. anonymous
    • 5 years ago
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    oh..i made a mistake, the axis of hyperbola is y axis

  10. anonymous
    • 5 years ago
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    right ishan, n then 25 comes in the denominator of 400 :)

  11. anonymous
    • 5 years ago
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    So am i still right then? y=+-4/5x and (0,+-4)

  12. anonymous
    • 5 years ago
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    So am i still right then? y=+-4/5x and (0,+-4)

  13. anonymous
    • 5 years ago
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    right :)

  14. anonymous
    • 5 years ago
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    sweet thanks

  15. anonymous
    • 5 years ago
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    54 to99 simplest form

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