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anonymous

  • one year ago

Find an equation in standard form for the hyperbola with vertices at (0, ±6) and foci at (0, ±9).

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  1. anonymous
    • one year ago
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    @phi

  2. phi
    • one year ago
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    the center is the average of the vertices

  3. anonymous
    • one year ago
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    So (0,0)

  4. phi
    • one year ago
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    the vertices are on the y-axis, so the hyperbola looks like a frown/smile combo that means the "y' goes first in the standard equation \[ \frac{(y-k)^2}{a^2}- \frac{(x-h)^2}{b^2}=1\] a is the distance from the center to the focus "c" is the distance from the center to the vertex we use a and c to find b a^2 + b^2 = c^2 we already know (h,k) is (0,0)

  5. phi
    • one year ago
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    oops, got that swapped: a is the distance from the center to the vertex "c" is the distance from the center to the focus

  6. anonymous
    • one year ago
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    alright.

  7. anonymous
    • one year ago
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    so a is 6 and c is 9

  8. phi
    • one year ago
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    yes, but in the equation you use a^2 = 36 and c^2=81 what is b^2 ?

  9. anonymous
    • one year ago
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    \[3\sqrt{13}\]

  10. anonymous
    • one year ago
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    b^2 is 117

  11. phi
    • one year ago
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    a^2 + b^2 = c^2 36 + b^2 = 81

  12. anonymous
    • one year ago
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    oh my bad

  13. anonymous
    • one year ago
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    b^2 = 45 b=3sqrt5

  14. phi
    • one year ago
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    now fill in the numbers to get the equation

  15. anonymous
    • one year ago
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    thanks. I still have a few more

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