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anonymous

  • one year ago

Regan is trying to find the equation of a quadratic that has a focus of (−2, 5) and a directrix of y = 13. Describe to Regan your preferred method for deriving the equation. Make sure you use Regan's situation as a model to help her understand.

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  1. anonymous
    • one year ago
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    I don't quite understand how to find a when writing the equation.. that's all I need help with.

  2. anonymous
    • one year ago
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    @jchick

  3. jchick
    • one year ago
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    Sorry it took so long to reply lol

  4. anonymous
    • one year ago
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    its cool XD

  5. anonymous
    • one year ago
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    I waited till I saw you weren't actively helping someone

  6. jchick
    • one year ago
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    The definition of a parabola is that the distance of any point on it from the focus is equal to the distance of the point from the directrix. So if (x,y) is a point on the parabola, then find the distance of (x,y) from the focus (-2,5). Find the distance of (x,y) from y = 13. Equate them and you will have your quadratic equation.

  7. jchick
    • one year ago
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    (x+2)^2+(y−5)^2=(y−13)^2 Simplify above.

  8. anonymous
    • one year ago
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    but how do you find (x,y)?

  9. jchick
    • one year ago
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    no matter what value x is, y will always be 13

  10. anonymous
    • one year ago
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    why?

  11. jchick
    • one year ago
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    directrix y = 13 can be represented as the point (x, 13)

  12. anonymous
    • one year ago
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    oh okay, that makes sense.

  13. jchick
    • one year ago
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    So after inputting the points, you have: (x−(−2))2+(y−5)2−−−−−−−−−−−−−−−−−−√=(x−x)2+(y−13)2−−−−−−−−−−−−−−−−√

  14. jchick
    • one year ago
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    You can then square both sides to get: (x−(−2))2+(y−5)2= (x−x)2+(y−13)2

  15. jchick
    • one year ago
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    Does this make sense so far?

  16. anonymous
    • one year ago
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    im not sure... itdoes a little bit

  17. jchick
    • one year ago
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    Ok where do you need explanation?

  18. jchick
    • one year ago
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    You can then square both sides to get: (x−(−2))^2+(y−5)^2= (x−x)^2+(y−13)^2

  19. jchick
    • one year ago
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    sorry that is the correct version

  20. anonymous
    • one year ago
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    what are you plugging them into?

  21. jchick
    • one year ago
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    What do you mean?

  22. jchick
    • one year ago
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    Where?

  23. anonymous
    • one year ago
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    like are you plugging those into an equation??

  24. jchick
    • one year ago
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    And then that simplifies to (x+2)^2+(y−5)^2=(y−13)^2

  25. jchick
    • one year ago
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    Yes

  26. anonymous
    • one year ago
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    okay, what equation?

  27. jchick
    • one year ago
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    (x−(−2))2+(y−5)2−−−−−−−−−−−−−−−−−−√= (x−x)2+(y−13)2−−−−−−−−−−−−−−−−√

  28. anonymous
    • one year ago
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    im so confused....

  29. jchick
    • one year ago
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    inserted points (-2,5) and (x, 13) the distance formula: (x−x1)2+(y−y2)2−−−−−−−−−−−−−−−−−√=(x−x2)2+(y−y2)2−−−−−−−−−−−−−−−−−√

  30. jchick
    • one year ago
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    Do you understand now?

  31. anonymous
    • one year ago
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    like I was learning it like this... The graph of the quadratic is a parabola. The directrix is horizontal, so the parabola is vertical. The focus lies below the directrix, so the parabola opens downwards. General equation of a down-opening parabola:    y = a(x - h)² + k with    a < 0    vertex (h, k)    focal length p = | 1/(4a) |    focus (h, k-p)    directrix y = k+p Plug your data into the equations and solve for a now im really confused... when I was only confused about how to figure out what a is XD

  32. jchick
    • one year ago
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    Oh ok

  33. jchick
    • one year ago
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    Sorry for confusion

  34. anonymous
    • one year ago
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    its cool I just didn't know what you were plugging it all into.. XD

  35. jchick
    • one year ago
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    Having internet trouble please hang on a second

  36. anonymous
    • one year ago
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    okay

  37. anonymous
    • one year ago
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    I've never seen this done this way. Pretty cool @jchick It looks like this is using the definition of a parabola as the set of points equidistant between the focus and directrix. A circle is a set of point equidistant from the same point. The equation for a circle is \[(x-h)^2+(y-k)^2=r^2\]. jchick set the equations for 2 circles equal to each other, one for the focus with (-2, 5) and the other for the directrix (x, 13). That's where this came from \[(x+2)^2+(y-5)^2=(x-x)^2+(y-13)\] If you mulitply everything out, and then complete the square you can find a with this method

  38. anonymous
    • one year ago
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    so do you use the distance formula to find a?

  39. anonymous
    • one year ago
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    that's pretty much what it works out to being. and you don't actually need to complete the square. a will be the coefficient of x²

  40. jchick
    • one year ago
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    Sorry I am back and thank you @peachpi

  41. anonymous
    • one year ago
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    just FOIL out the equation and you'll have an expanded equation for the parabola

  42. anonymous
    • one year ago
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    im still really confused... what is r in that equation??

  43. anonymous
    • one year ago
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    Maybe try a different method since this is confusing. We know that the vertex is halfway between the focus and directrix, so it's (-2, 9) and the parabola opens down like you said above. So the focal length is 9 - 5 = 4 |dw:1439944412409:dw|

  44. anonymous
    • one year ago
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    how do you find the vertex?

  45. jchick
    • one year ago
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    (x-a)^2 =-4p(y-b) vertex =(a,b)

  46. jchick
    • one year ago
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    @kaylamarie79 do you remember the standard form?

  47. anonymous
    • one year ago
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    The parabola opens down so the vertex has to be directly above the focus, meaning the x-coordinate has to be -2. The vertex is halfway between focus and directrix, so it's halfway between 5 and 13, so the y-coordinate has to be 9. Vertex = (-2, 9). To find a: From your list of equations above, \[p=\left| \frac{ 1 }{ 4a } \right|\] Since you know p, you can solve for a.

  48. anonymous
    • one year ago
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    y=ax^2+bx+c

  49. anonymous
    • one year ago
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    @peachpi I dont know p

  50. anonymous
    • one year ago
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    p is the focal distance

  51. anonymous
    • one year ago
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    p = distance from vertex to focus = distance from vertex to directrix = half the distance from focus to directrix

  52. anonymous
    • one year ago
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    1 min im trying wrap my mind around this..

  53. anonymous
    • one year ago
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    what are you thinking?

  54. anonymous
    • one year ago
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    all I can think is i'm sooo confused... like I don't understand where your getting a the focal point(p) or the vertex... and I keep getting more confused as y'all repeat things.. like i'm terrible in math and have to see every single step or the entire equation gets jumbled around in my head...

  55. anonymous
    • one year ago
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    ok. starting from the beginning. The directrix is y = 13. The focus is (-2, 5).|dw:1439945735884:dw|

  56. anonymous
    • one year ago
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    uh- huh

  57. anonymous
    • one year ago
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    The vertex has to be halfway between them. 9 is halfway between 5 and 13, so the vertex has to be (-2, 9)|dw:1439945891758:dw|

  58. anonymous
    • one year ago
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    how do you get the vertex point?

  59. anonymous
    • one year ago
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    what number is halfway between 5 and 13?

  60. anonymous
    • one year ago
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    oh okay. how do you know the -2 though?

  61. anonymous
    • one year ago
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    because the parabola opens down, the focus and vertex must have the same x-coordinate

  62. anonymous
    • one year ago
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    is it always equal as the focus or only when opening down?

  63. anonymous
    • one year ago
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    only when opening up or down. When it opens left or right the y-coordinates are the same

  64. anonymous
    • one year ago
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    oh okay that makes sense

  65. anonymous
    • one year ago
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    that's the vertex. Now for p. That's the distance between the focus and vertex, so it's p = 9 - 5.

  66. anonymous
    • one year ago
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    p = 4

  67. anonymous
    • one year ago
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    with me?

  68. anonymous
    • one year ago
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    yeah, so to get a you use the focal length equation but just plug in p?

  69. anonymous
    • one year ago
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    yes. but since it opens down make p be -4 and you can drop the absolute value signs

  70. anonymous
    • one year ago
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    ohhhhh see the other ways just got really confusing..

  71. anonymous
    • one year ago
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    okk cool

  72. jchick
    • one year ago
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    Did you find your answer?

  73. anonymous
    • one year ago
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    yeah, I did.., I understand now XD

  74. jchick
    • one year ago
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    Ok good

  75. anonymous
    • one year ago
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    thank you guys for helping

  76. jchick
    • one year ago
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    Yeah no problem and sorry about being confusing lol

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