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Android

  • 5 years ago

What is standard form of this hyperbola? 9y^2 - 16x^2 - 144 = 0 Then I can graph it.

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  1. anonymous
    • 5 years ago
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    Hyperbolas are of the form\[(x-h)^{2}/a +(y-k)^{2}/b=1\]

  2. android
    • 5 years ago
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    How do I convert it to that?

  3. anonymous
    • 5 years ago
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    Use intuition, from your equation how can you get a one on the right and at the same time some number under x and y?

  4. android
    • 5 years ago
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    Add 144 to the 0?

  5. anonymous
    • 5 years ago
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    Great, you're on the right track, continue.

  6. android
    • 5 years ago
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    Would I now divide the 9 and 16 by 144?

  7. anonymous
    • 5 years ago
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    Yes

  8. android
    • 5 years ago
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    Is that the final answer? 9y^2/144 - 16x^2/144 = 1

  9. anonymous
    • 5 years ago
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    9 can divide 144, and 16 into 144

  10. android
    • 5 years ago
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    Divide 9 by 44?

  11. anonymous
    • 5 years ago
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    You want a^2 and b^2 at the bottom, so divide 144 by 9 and so on

  12. android
    • 5 years ago
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    9y^2/16 - 16x^2/12 = 1

  13. anonymous
    • 5 years ago
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    You're on the right track, you should cancel the 9 and 16 and calculations off a little on b\[(y-0)^{2}/4^{2} -(x-0)^{2}/3^{2}=1\]Good job

  14. android
    • 5 years ago
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    Thanks

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