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anonymous

  • 5 years ago

Find the vertices, foci, and asymptotes of the hyperbola. x^2 − 7y^2 = 8 can any one help me with this?

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  1. myininaya
    • 5 years ago
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    does this help? you can write the equation as (x-0)^2/8-(y-0)^2/(8/7)=1

  2. anonymous
    • 5 years ago
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    yes iv got all that. im just having trouble finding the asymptotes.

  3. myininaya
    • 5 years ago
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    it says do sqrt(8)/sqrt(8/7) is the slope of the asymptote

  4. myininaya
    • 5 years ago
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    =sqrt(7)

  5. anonymous
    • 5 years ago
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    yes ive done that but i get a weird radical that i dont know how to reduce?...

  6. myininaya
    • 5 years ago
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    what is it?

  7. anonymous
    • 5 years ago
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    sqrt 65/7 / 2 sqrt 2

  8. myininaya
    • 5 years ago
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    sqrt(65/7)/[2sqrt(2)]?

  9. anonymous
    • 5 years ago
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    yes

  10. myininaya
    • 5 years ago
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    thats not the slope is it? what is that?

  11. anonymous
    • 5 years ago
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    the asymptotes

  12. myininaya
    • 5 years ago
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    I don't know about that. i have to go though. i'm sorry. if you want to simplify the above that you wrote, it is sqrt(65)/[2*sqrt(2)]

  13. anonymous
    • 5 years ago
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    thanks.

  14. anonymous
    • 5 years ago
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    sqrt(65)/[2*sqrt(2)]???

  15. myininaya
    • 5 years ago
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    oops lol

  16. myininaya
    • 5 years ago
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    sorry it was suppose to be 1/2 * sqrt(65/14)

  17. anonymous
    • 5 years ago
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    ok let me try it.

  18. myininaya
    • 5 years ago
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    you have double fraction so you change to multiplication by flipping the second fraction

  19. anonymous
    • 5 years ago
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    1/2 * sqrt(65/14) is the final answer right?

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