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anonymous

  • one year ago

Anyone here good with ellipses? If so, please help!! Find the center, vertices, and foci of the ellipse with equation x squared divided by four hundred plus y squared divided by two hundred and fifty six = 1 A. Center: (0, 0); Vertices: (-20, 0), (20, 0); Foci: (-12,0), (12, 0) B. Center: (0, 0); Vertices: (-20, 0), (20, 0); Foci: (-16,0), (16, 0) C. Center: (0, 0); Vertices: (0, -20), (0, 20); Foci: (0,-16), (0, 16) D. Center: (0, 0); Vertices: (0, -20), (0, 20); Foci: (0,-12), (0, 12) I know the answer is either B or C

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  1. anonymous
    • one year ago
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    And also for Find an equation in standard form for the ellipse with the vertical major axis of length 6, and minor axis of length 4. I am pretty sure the answer is: x^2/9 + y^2/4 = 1 Is that correct?

  2. Nnesha
    • one year ago
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    I have few minutes lets see if i can help u :=)

  3. Nnesha
    • one year ago
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    1st) is it vertical or horizontal ?

  4. anonymous
    • one year ago
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    Ok thank you! And for the first one it is horizontal. @Nnesha

  5. Nnesha
    • one year ago
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    yes right so formula for foci \[\huge\rm (h \pm c , k)\]

  6. Nnesha
    • one year ago
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    first of all you need to find c

  7. Nnesha
    • one year ago
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    use this equation \[\huge\rm c^2=a^2-b^2\]

  8. anonymous
    • one year ago
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    Would c = SqRt656? @Nnesha

  9. Nnesha
    • one year ago
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    remember (h,k) is the center point it's horizontal so only x values would change y would stay same no change in y values

  10. Nnesha
    • one year ago
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    no what is a^2 and b^2 ?

  11. Nnesha
    • one year ago
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    \[\huge\rm \frac{ (x-h)^2 }{ a^2 }+\frac{ (y-k)^2 }{ b^2 }\] standard form of ellipse

  12. anonymous
    • one year ago
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    Wouldn't a^2 = 256 and b^2 = 400?

  13. anonymous
    • one year ago
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    Oh wait.

  14. anonymous
    • one year ago
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    12 = c?

  15. Nnesha
    • one year ago
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    okay for ellipse a= bigger number if a is under the x then it's horizontal and if a is under y then it would be vertical

  16. Nnesha
    • one year ago
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    yes !

  17. Nnesha
    • one year ago
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    so now what are the foci points ?

  18. anonymous
    • one year ago
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    +/12, 0?

  19. anonymous
    • one year ago
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    Forgot the ( ) sorry

  20. Nnesha
    • one year ago
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    \[(\pm 12,0)\] looks great should be plus minus sign

  21. anonymous
    • one year ago
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    Yes that's what I meant, sorry.

  22. anonymous
    • one year ago
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    So my answer would actually be A. Ok, thanks! Could you help me with the 2nd one?

  23. anonymous
    • one year ago
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    And also with this one: The rectangular piece of property measures 10 mi by 4 mi. Find an equation for the ellipse if the path is to touch the center of the property line on all 4 sides. I believe my answer would be: x^2/25 + y^2/4 = 1 But I just wanted to make sure.

  24. Nnesha
    • one year ago
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    \(\color{blue}{\text{Originally Posted by}}\) @jdoherty And also for Find an equation in standard form for the ellipse with the vertical major axis of length 6, and minor axis of length 4. I am pretty sure the answer is: x^2/9 + y^2/4 = 1 Is that correct? \(\color{blue}{\text{End of Quote}}\) *vertical*

  25. anonymous
    • one year ago
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    ? @Nnesha

  26. Nnesha
    • one year ago
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    if ellipse is vertical then bigger number supposed to be under x or y ?

  27. anonymous
    • one year ago
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    Y?

  28. Nnesha
    • one year ago
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    yes right try to draw points on the paper

  29. Nnesha
    • one year ago
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    gtg sorry. :(

  30. anonymous
    • one year ago
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    Ok :( Are you free at all tomorrow?

  31. anonymous
    • one year ago
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    @saseal Could you help with the other two questions?

  32. anonymous
    • one year ago
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    ok

  33. anonymous
    • one year ago
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    Thank you! @saseal

  34. anonymous
    • one year ago
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    1. The rectangular piece of property measures 10 mi by 4 mi. Find an equation for the ellipse if the path is to touch the center of the property line on all 4 sides. I believe my answer would be: x^2/25 + y^2/4 = 1 But I just wanted to make sure. 2. Find an equation in standard form for the ellipse with the vertical major axis of length 6, and minor axis of length 4. I am pretty sure the answer is: x^2/9 + y^2/4 = 1 Is that correct?

  35. anonymous
    • one year ago
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    @saseal ^

  36. anonymous
    • one year ago
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    first one is correct...2nd one is debatable because the question is not very specified

  37. anonymous
    • one year ago
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    if they specifed the major axis is x and minor is y then you are correct

  38. anonymous
    • one year ago
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    Hmm...how could I determine for sure?

  39. anonymous
    • one year ago
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    hmm you are wrong...it says vertical major axis

  40. anonymous
    • one year ago
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    Oh ok. So that would make the y axis the major axis?

  41. anonymous
    • one year ago
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    Oh wait. would it be x^2/4 + y^2/9 = 1? @saseal

  42. anonymous
    • one year ago
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    yea

  43. anonymous
    • one year ago
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    Ok

  44. anonymous
    • one year ago
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    Could you help me a couple more questions? I kind of am just getting this stuff. Lol. @saseal

  45. anonymous
    • one year ago
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    ok

  46. anonymous
    • one year ago
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    Find the vertices and foci of the hyperbola with equation x squared over nine minus y squared over sixteen = 1 This is for Hyperbolas...

  47. anonymous
    • one year ago
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    \[\frac{ x^2 }{ 3^2 } - \frac{ y^2 }{ 4^2 } = 1\]

  48. anonymous
    • one year ago
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    now it becomes y^2/4^2 - x^2/3^2

  49. anonymous
    • one year ago
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    \[\frac{ y^2 }{ 4^2 } - \frac{ x^2 }{ 3^2 } =1\]

  50. anonymous
    • one year ago
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    Ok so then what can I do?

  51. anonymous
    • one year ago
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    you know where thid hyperbola opens open?

  52. anonymous
    • one year ago
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    Not really like when it's graphed?

  53. anonymous
    • one year ago
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    |dw:1439004618140:dw|

  54. anonymous
    • one year ago
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    Ok so they open up where the two points are? Like the end points of the opening?

  55. anonymous
    • one year ago
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    kinda...abit like parabola x2

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