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anonymous

  • one year ago

HELP!!!! Find the foci of 16y^2-36x^2-576=0. Then graph the hyperbole.

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  1. Loser66
    • one year ago
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    move -576 to the right side, what do you get?

  2. anonymous
    • one year ago
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    Positive 576

  3. Loser66
    • one year ago
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    yes, now, divide both sides by 576, what do you get?

  4. anonymous
    • one year ago
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    1/36y^2-1/16x^2

  5. Loser66
    • one year ago
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    =??

  6. anonymous
    • one year ago
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    =1?

  7. Loser66
    • one year ago
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    \[\dfrac{y^2}{6^2}-\dfrac{x^2}{4^2}=1\] right?

  8. anonymous
    • one year ago
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    yeah

  9. Loser66
    • one year ago
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    Hence, its b =6, a =4 ok?

  10. anonymous
    • one year ago
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    yes

  11. Loser66
    • one year ago
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    foci c, \(c=\sqrt{a^2+b^2}=??\)

  12. anonymous
    • one year ago
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    thanks can you stay with me for just a second i need to write this down

  13. anonymous
    • one year ago
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    ok is it square root of 52

  14. Loser66
    • one year ago
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    You need know where c is. Your equation has y first, hence its graph is vertical, right? Hence foci will be (0, -c) and (0,c) . Yes for c =sqrt 52

  15. Loser66
    • one year ago
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    |dw:1437522466412:dw|

  16. anonymous
    • one year ago
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    do you put the foci into decimal form for the graph or do you just leave it as c=sqrt52

  17. Loser66
    • one year ago
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    |dw:1437522532448:dw|

  18. Loser66
    • one year ago
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    leave it as it is.

  19. anonymous
    • one year ago
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    okay so I would just put the sqrt52 into the graph vertically and be done with the graphing

  20. anonymous
    • one year ago
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    okay im gonna assume its done

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