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sloppycanada

  • one year ago

Ellipses http://gyazo.com/e897fc96e592ab1e5dcba825538b6d7e y^2/100 - x^2/16 = 1

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  1. anonymous
    • one year ago
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    no not this time

  2. anonymous
    • one year ago
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    center is not \((0,0)\) find the center first

  3. sloppycanada
    • one year ago
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    (1,2)?

  4. anonymous
    • one year ago
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    yes

  5. anonymous
    • one year ago
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    so it is going to look like \[\frac{(x-1)^2}{a^2}-\frac{(y-2)^2}{b^2}=1\]

  6. anonymous
    • one year ago
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    you need \(a\) and \(b\)

  7. sloppycanada
    • one year ago
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    But wouldn't x and y be switched? Since the graph is going a different way?

  8. anonymous
    • one year ago
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    no

  9. sloppycanada
    • one year ago
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    So it's not like an ellipses that way.

  10. anonymous
    • one year ago
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    |dw:1435717476572:dw|

  11. anonymous
    • one year ago
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    scuse the crappy pictures

  12. sloppycanada
    • one year ago
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    No worries, mine are worse Okay so for this graph it'd be - (x^2-2)/100 - (y^2-1)/16 = 1

  13. sloppycanada
    • one year ago
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    Also for this one - you said it was x than y. http://gyazo.com/99da097eae7203d795aaab65586d4f48

  14. anonymous
    • one year ago
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    then i made a mistake

  15. anonymous
    • one year ago
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    also no, not 100 and 16

  16. anonymous
    • one year ago
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    the distance from the center to the vertex is 5

  17. anonymous
    • one year ago
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    not the whole length from one vertex to the other, which is 10 so \(25\) goes under the \(x\) part

  18. anonymous
    • one year ago
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    similarly under the \(y\) part put a 4, not 16

  19. sloppycanada
    • one year ago
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    (x^2-2)/25 - (y^2-1)/4 = 1 is my final answer?

  20. anonymous
    • one year ago
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    yes want to check it?

  21. anonymous
    • one year ago
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    oh no !!

  22. sloppycanada
    • one year ago
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    No?

  23. anonymous
    • one year ago
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    not \(x^2-1\) up top but rather \((x-1)^2\)

  24. anonymous
    • one year ago
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    sorry \((x-2)^2\)

  25. sloppycanada
    • one year ago
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    (x-2)^2/25 - (y-1)^2/4 = 1

  26. anonymous
    • one year ago
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    here is the check that \[\frac{(x-2)^2}{25}-\frac{(y-1)^2}{4}=1\] is right http://www.wolframalpha.com/input/?i=hyperbola+%28x-1%29^2%2F25-%28y-2%29^2%2F4%3D1

  27. anonymous
    • one year ago
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    hard to tell from the graphs, but you can see from the vertices that it is correct

  28. sloppycanada
    • one year ago
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    Okay thanks!

  29. sloppycanada
    • one year ago
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    So many hyperbola questions...

  30. anonymous
    • one year ago
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    there is a certain sameness to them

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