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Babynini

  • one year ago

Ellipses help!

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  1. Babynini
    • one year ago
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  2. Babynini
    • one year ago
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    @ganeshie8 I'm not sure how to keep 1 on the right side while simplifying the left

  3. Babynini
    • one year ago
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    @freckles ? :)

  4. Nnesha
    • one year ago
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    it's not ellipse

  5. Nnesha
    • one year ago
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    http://prntscr.com/7cbqdf :-)

  6. Babynini
    • one year ago
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    Sorry, I meant hyperbola

  7. Nnesha
    • one year ago
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    so is it horizontal or vertical ?

  8. Nnesha
    • one year ago
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    do you know ?

  9. Babynini
    • one year ago
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    Sorry I didn't get your responses until now! grr pc

  10. Babynini
    • one year ago
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    horizontal?

  11. Babynini
    • one year ago
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    \[\frac{ x^2 }{ \frac{ 1 }{ 36 } }-\frac{ y^2 }{ \frac{ 1 }{ 64 } }=1\]

  12. Babynini
    • one year ago
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    a= 1/6 b=1/8 c=5/24

  13. Babynini
    • one year ago
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    yeah?

  14. Nnesha
    • one year ago
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    it's hyperbola so its not depends on a(bigger number) if x comes first then it's horizontal and if y comes first then it's vertical \[x^2 - y^2 = 1 \]^^^ horizontal \[y^2 - x^2 =1\] vertical ^

  15. Babynini
    • one year ago
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    ooo I didn't catch that in class. That is helpful ^-^ so yeah it's horizontal.

  16. Nnesha
    • one year ago
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    yep right

  17. Babynini
    • one year ago
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    do you know if the a b and c are right?

  18. Nnesha
    • one year ago
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    gimme a sec

  19. Nnesha
    • one year ago
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    alright

  20. Babynini
    • one year ago
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    ok :) I'm at school and going to drive home so brb in like 15 :)

  21. Nnesha
    • one year ago
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    oh okay

  22. Babynini
    • one year ago
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    ...so that was waay more than 15, sorry! back now.

  23. Babynini
    • one year ago
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    @Nnesha

  24. Nnesha
    • one year ago
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    alright to find c u have to use this formula \[\huge\rm c^2 = a^2 + b^2\] plug in a^2 and b^2 value solve for c

  25. Babynini
    • one year ago
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    Earlier I posted what I got for a, b, and c :)

  26. Babynini
    • one year ago
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    a= 1/6 b=1/8 c=5/24

  27. Nnesha
    • one year ago
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    ohh nvm i forgot

  28. Nnesha
    • one year ago
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    that's right now bec it's horizontal formula for focus \[\huge\rm (h \pm c ,k)\] (h,k) is the center point add c value in x coordinate

  29. Nnesha
    • one year ago
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    center is ( 0 ,0 ) in that question

  30. Nnesha
    • one year ago
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    so focus is = ?

  31. Babynini
    • one year ago
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    (0, plus or minus 5/24)

  32. Nnesha
    • one year ago
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    x-coordinate not y it's horizontal so add h +/- c

  33. Nnesha
    • one year ago
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    \[\huge\rm (h \pm \color{red}{c} ,k)\] (h,k) is the center point h + c and h - c

  34. Nnesha
    • one year ago
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    alright seems like u r afk so for vertex \[\huge\rm (h \pm a ,k)\] and for asy \[\huge\rm y = k \pm \frac{ b }{ a }(x-h)\] (h.k) is the center

  35. Nnesha
    • one year ago
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    pm means \[\huge\rm (h + a , k) and ( h -a ,k)\]

  36. Babynini
    • one year ago
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    did you get my msg?

  37. Babynini
    • one year ago
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    wait what? focal point is... (0,a) then?

  38. Nnesha
    • one year ago
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    nope

  39. Nnesha
    • one year ago
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    wat msg ?

  40. Babynini
    • one year ago
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    I sent you a message on here. Or i thought i did

  41. Babynini
    • one year ago
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    (0,c) ?

  42. Babynini
    • one year ago
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    what is the h?

  43. Nnesha
    • one year ago
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    (h , k) is the center which is (0,0) in this question did u read my comments ? :-)

  44. Babynini
    • one year ago
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    Yeah, but if it's that then the foci are (0, pm 5/24)

  45. Nnesha
    • one year ago
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    nope add into x-coordinate not y \[\huge\rm (h \pm \color{red}{c} ,k)\] h + c and h - c

  46. Babynini
    • one year ago
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    I don't understand that D: c - 5/24

  47. Babynini
    • one year ago
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    * c = 5/24

  48. Nnesha
    • one year ago
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    what is h and k ?

  49. Babynini
    • one year ago
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    (0,0)

  50. Nnesha
    • one year ago
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    yep right nice face! 0.0 alright so it should be like this \[(0 \pm \frac{ 5 }{ 24} , 0)\]

  51. Babynini
    • one year ago
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    why is the k there? The previous problems I've done never have had that

  52. Nnesha
    • one year ago
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    what k ?

  53. Babynini
    • one year ago
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    After the c :P

  54. Nnesha
    • one year ago
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    k = 0 y- coordinate now bec it's horizontal u have to add into x-coordinate not y so it's yep right nice face! 0.0 alright so it should be like this \[( \pm \frac{ 5 }{ 24} , 0)\]

  55. Babynini
    • one year ago
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    so the final answer should be (0, pm5/24, 0)

  56. Babynini
    • one year ago
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    vertex is (0, pm1/6, 0)

  57. Babynini
    • one year ago
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    asy y=0 pm6/8(x-0)

  58. Nnesha
    • one year ago
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    \(\color{blue}{\text{Originally Posted by}}\) @Babynini so the final answer should be (0, pm5/24, 0) \(\color{blue}{\text{End of Quote}}\) why two zeros ? \[(0 \pm \frac{ 5 }{ 24} , 0)\] \[0 + \frac{ 5 }{ 24 } =? ?\] \[0-\frac{ 5 }{ 24 }=?\]

  59. Babynini
    • one year ago
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    ..oh xD foci: -5/24, 0 5/24, 0

  60. Nnesha
    • one year ago
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    yes right

  61. Babynini
    • one year ago
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    asy y=pm6/8(x-0)

  62. Nnesha
    • one year ago
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    yep solve that reduce the fraction

  63. Babynini
    • one year ago
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    or just pm6/8x

  64. Nnesha
    • one year ago
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    reduce the fraction

  65. Nnesha
    • one year ago
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    6/8= ?

  66. Babynini
    • one year ago
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    xD 3/4

  67. Nnesha
    • one year ago
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    yep right so y = 3/4x

  68. Babynini
    • one year ago
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    okies :)

  69. Nnesha
    • one year ago
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    done!?

  70. Babynini
    • one year ago
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    it's all green checks! :) thanks so much

  71. Nnesha
    • one year ago
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    np :-)

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