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anonymous

  • one year ago

Find the standard form of the equation of the parabola with a focus at (-4, 0) and a directrix at x = 4. Could someone please help?

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  1. wintersuntime
    • one year ago
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    Write an equation for the parabola with focus (4,0) and directrix y=2. Thanks! .. Standard form of parabola: (x-h)^2=4p(y-k), with (h,k) being the (x,y) coordinates of the vertex Given parabola opens downward with axis of symmetry at x=4. Vertex at (4,1). p=1 Equation of parabola: (x-4)^2=-4(y-1) (ans) see the graph below as a visual check on the answer .. y=1-(x-4)^2/4

  2. anonymous
    • one year ago
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    I'm so confused I'm sorry @wintersuntime

  3. wintersuntime
    • one year ago
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    Are you trying to graph it ?

  4. wintersuntime
    • one year ago
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    And it's okay I love helping others

  5. anonymous
    • one year ago
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    No. These are the answer I am given: A. y^2 = -8x B. 16y = x^2 C. y = -1/16x^2 D. x = -1/16y^2

  6. wintersuntime
    • one year ago
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    What do you think the answer will be ?

  7. anonymous
    • one year ago
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    I am thinking C. @wintersuntime

  8. wintersuntime
    • one year ago
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    why ?

  9. anonymous
    • one year ago
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    Honestly, I'm not sure. I just thought that standard form looked most similar to that choice. @wintersuntime

  10. anonymous
    • one year ago
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    its not C

  11. anonymous
    • one year ago
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    Could you explain how to find the correct answer then? I am so lost. @saseal

  12. anonymous
    • one year ago
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    |dw:1438996328137:dw|

  13. anonymous
    • one year ago
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    the parabola is opening down because the focus is below the directrix

  14. anonymous
    • one year ago
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    But the parabola's focus is at -4, 0. So would that mean the parabola's focus would be shifted to the left 4? Not down 4?

  15. anonymous
    • one year ago
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    no, focus has nothing to do with shifting

  16. anonymous
    • one year ago
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    Yes, but you put the point at (0, -4) it looks like.

  17. anonymous
    • one year ago
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    lel i made a mistake at the drawing suppose to be (0,p)

  18. anonymous
    • one year ago
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    Ok, but I'm still confused as to what the standard form would look like?

  19. anonymous
    • one year ago
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    the standard form flavor looks like this \[x^2=4py\] or \[y^2=4px\]

  20. anonymous
    • one year ago
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    depends on where your parabola opens out

  21. anonymous
    • one year ago
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    Ok so since the parabola opens down, would it be y^2 = 4px?

  22. anonymous
    • one year ago
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    |dw:1438996750393:dw|

  23. anonymous
    • one year ago
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    |dw:1438996883193:dw|

  24. anonymous
    • one year ago
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    so its not y^2

  25. anonymous
    • one year ago
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    Oh ok, so it would be in the form of x^2 = 4py?

  26. anonymous
    • one year ago
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    yes

  27. anonymous
    • one year ago
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    So the answer then would be B? @saseal

  28. Loser66
    • one year ago
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    Tip: no need to consider whether p >0 or p <0, if you see the focus, just plug that p in For example, on your problem, focus is (0,-4) , hence plug -4 in the form x^2 = 4 p y x^2 = 4*(-4) y.

  29. Loser66
    • one year ago
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    then isolate y to get the form.

  30. anonymous
    • one year ago
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    should be x^2 = -16y

  31. anonymous
    • one year ago
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    That's not the answer. I just checked.

  32. anonymous
    • one year ago
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    I think it's y = -1/16x^2

  33. Loser66
    • one year ago
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    surely not \(x^2 = -16y\) divided both sides by -16, you get y = -(1/16)x^2

  34. anonymous
    • one year ago
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    Ok I was correct. Thank you!

  35. anonymous
    • one year ago
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    I wish I could give you all medals.

  36. Loser66
    • one year ago
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    give it to saseal, I have more than 3 thousands of it. hehehe.

  37. anonymous
    • one year ago
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    lol

  38. anonymous
    • one year ago
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    Could you guys just check a couple for me? I'm really confused on these, and have them done, just want to make sure they're correct.

  39. anonymous
    • one year ago
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    alright

  40. anonymous
    • one year ago
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    Find the vertex, focus, directrix, and focal width of the parabola. x = 3^y2 Would the answer be: Vertex: (0,0) Focus: (1/12,0) Focus Width: .333 or 1/12 Directrix: -1/12

  41. anonymous
    • one year ago
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    \[x=3y^2\]\[y^2=\frac{ 1 }{ 3 }x\]\[y^2 = 4(\frac{ 1 }{ 12 }) x\]

  42. anonymous
    • one year ago
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    Ok...? So I'm incorrect?

  43. anonymous
    • one year ago
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    focus width is wrong and directrix usually they write it as directrix = x = -1/12 so people know which axis it is on

  44. anonymous
    • one year ago
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    Oh ok sorry. But my choice for the vertex, focus, and directrix to all remain the same, the focal width has to be .33

  45. anonymous
    • one year ago
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    yes

  46. anonymous
    • one year ago
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    focal width aka latus rectum = 4p

  47. anonymous
    • one year ago
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    That's not one of my choices. I think they wanted you to solve for the width.

  48. anonymous
    • one year ago
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    |dw:1438997923182:dw|

  49. anonymous
    • one year ago
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    Ok thanks!

  50. anonymous
    • one year ago
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    For: A building has an entry the shape of a parabolic arch 74 ft high and 28 ft wide at the base as shown below. Would my answer be: y = −37/98x^2

  51. anonymous
    • one year ago
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    |dw:1438998052704:dw|

  52. anonymous
    • one year ago
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    Ok

  53. anonymous
    • one year ago
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    so its definitely p<0

  54. anonymous
    • one year ago
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    So what would that mean for the ending answer?

  55. Loser66
    • one year ago
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    To this kind of problem, you need derive from the standard form Vertex (0,74), Two points are (-14,0) and (14,0) ok?

  56. anonymous
    • one year ago
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    Ok

  57. Loser66
    • one year ago
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    Vertex form is \(y = a(x-h)^2 +k\) \(y = ax^2 + 74\)

  58. anonymous
    • one year ago
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    |dw:1438998433802:dw|

  59. anonymous
    • one year ago
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    always better t o draw it out haha

  60. Loser66
    • one year ago
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    Now, pick one of the point, I pick (14,0) , that is x =14 , y =0 and plug them in \(0= a (14)^2 +74\) \(a(14)^2 = -74\) \(a = \dfrac{-74}{196}\)

  61. Loser66
    • one year ago
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    Now, replace back \(y = \dfrac{-74}{196}x^2\) @saseal they didn't say the base of the parabola is the line passed through focus. You cannot apply the form of focus, directrix here.

  62. Loser66
    • one year ago
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    @jdoherty Got what I meant?

  63. anonymous
    • one year ago
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    Ok, so my answer would be correct then. I do understand. (:

  64. anonymous
    • one year ago
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    yea ikr thats why i didnt use anything to do with focus

  65. Loser66
    • one year ago
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    your answer is (-37/98)x^2, that is not correct.

  66. anonymous
    • one year ago
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    But I just simplified the fraction, dividing both the top and bottom by 2... And I just have one more question. Find the standard form of the equation of the parabola with a vertex at the origin and a focus at (0, -7). Those still are the most confusing to me.

  67. anonymous
    • one year ago
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    x^2 = 4 (-7)y

  68. Loser66
    • one year ago
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    When they give you the focus or directrix, they want you to define the form of the parabola. You know that the focus is ALWAYS inside of the parabola, right? hence ,your parabola is up or down?

  69. anonymous
    • one year ago
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    So it would be: y = -1/28 x^2?

  70. anonymous
    • one year ago
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    @jdoherty refer to the drawings above to see where the parabola opens up

  71. anonymous
    • one year ago
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    I did refer to that, but based on the equation, I got my answer of y = -1/28x^2

  72. anonymous
    • one year ago
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    @saseal

  73. anonymous
    • one year ago
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    yea?

  74. anonymous
    • one year ago
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    so thats correct

  75. anonymous
    • one year ago
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    sorry i fell asleep

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