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anonymous

  • 5 years ago

Find the standard form of the equation of the hyperbola with the given characteristics. vertices: (2, - 6), (2, - 2) foci: (2, - 8), (2, 0) I know this much so far: (y+4)^2 (x-2)^2 ------ - ------ ? ?

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  1. amistre64
    • 5 years ago
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    the y is the heavy side; so this opens up and down

  2. amistre64
    • 5 years ago
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    teh distance between the foci is 8; so each foci is 4 from center; that 4 is either a or c :)

  3. anonymous
    • 5 years ago
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    ok so a

  4. amistre64
    • 5 years ago
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    we have 4 between the vertices; so each vertex is 2 from center

  5. anonymous
    • 5 years ago
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    right how do I find b^2 then?

  6. amistre64
    • 5 years ago
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    under the y we have 2^2 right? let me look this up some more :)

  7. anonymous
    • 5 years ago
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    ok I think actually I have to know what c^2 is cause C^2=a^2+b^2

  8. anonymous
    • 5 years ago
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    I'm thinkin' b^2=16...?

  9. amistre64
    • 5 years ago
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    tell me your understanding of a b and c; what do they measure?

  10. anonymous
    • 5 years ago
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    the foci are c units from the center the vertices are a units from the center

  11. amistre64
    • 5 years ago
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    so; a^2 + b^2 = c^2 then right?

  12. amistre64
    • 5 years ago
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    2 + b^2 = 64 b^2 = 62 then

  13. anonymous
    • 5 years ago
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    I think so....but that doesn't eem right maybe it's minus?

  14. anonymous
    • 5 years ago
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    my options for b^2 are 12 or 16 here but I don't see how they get either of those

  15. amistre64
    • 5 years ago
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    like this right?

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  16. amistre64
    • 5 years ago
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    i used 8 instead of 4 :)

  17. anonymous
    • 5 years ago
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    i think so...

  18. amistre64
    • 5 years ago
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    16-4 = 12

  19. amistre64
    • 5 years ago
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    b^2 = 12

  20. anonymous
    • 5 years ago
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    oh ok I was gonna say 16 cause 4^2=16 but 12 works too

  21. amistre64
    • 5 years ago
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    the minor axis is always smaller in number than the major axis

  22. amistre64
    • 5 years ago
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    since the major is the y on this one; and it is 16; the other has to be less

  23. amistre64
    • 5 years ago
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    or maybe i messed that up in thougth

  24. anonymous
    • 5 years ago
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    ugh... idk this is all a review for my final and I am thinking that so far i'm not doing too hot. YIKES

  25. amistre64
    • 5 years ago
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    y^2 x^2 --- - --- = 1 4 12 b^2 = focal distance(c)^2 - vertex distance(a)^2 in this case; the hyperbola; c is the greatest distance beacusae it is used as the hypotenust in the ellipse; a is the greatest distance becasue IT is used for ITS hypotenuse

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