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anonymous

  • one year ago

Medal & fan Using a directrix of y = -2 and a focus of (1, 6), what quadratic function is created?

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  1. anonymous
    • one year ago
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    @Nnesha @zepdrix @MeganEdward

  2. anonymous
    • one year ago
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    or @dan815 or @merchandize lol

  3. Nnesha
    • one year ago
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    |dw:1444448203481:dw| ;-;

  4. Jhannybean
    • one year ago
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    Recall that the vertex of a parabola lies halfway between the focus and the directrix

  5. Nnesha
    • one year ago
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    alright let's pick point (x,y) |dw:1444448374229:dw| we need to find distance between the point (x,y) and focus and the distance between point and directirx \[\sqrt{(x - a)^2+(y -b)^2}= \left| y-c \right|\] \[\sqrt{(x- 1)^2 + (y-6)^2} = (y-(-2))^2\]

  6. Nnesha
    • one year ago
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    npI y-c I is same as (y-c)^2 and (a,b) is the focus point any question ?

  7. anonymous
    • one year ago
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    Yes, I'm just completely lost :/ Don't understand this question as well as the other one.. I dont remember what a directrix even is

  8. Nnesha
    • one year ago
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    i see sorry directrix is a line y = c |dw:1444448897371:dw| and the focus point is at (1,6)

  9. dan815
    • one year ago
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    http://prntscr.com/8ppjc9

  10. Nnesha
    • one year ago
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    ahh distance between the point `on parabola ` i missd that part :(

  11. dan815
    • one year ago
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    http://prntscr.com/8ppjog

  12. anonymous
    • one year ago
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    I'm sorry, I just dont get it :(

  13. anonymous
    • one year ago
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    Ok mabey I can help

  14. Nnesha
    • one year ago
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    y = c ( it's horizontal line ) so y =-2 draw a horizontal line at -2 |dw:1444449356899:dw| red line is directrix right

  15. anonymous
    • one year ago
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    Right

  16. Nnesha
    • one year ago
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    now draw focus point (1,6) |dw:1444449417779:dw| read the definition on the dan's secreenshot :P |dw:1444449585947:dw|

  17. anonymous
    • one year ago
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    Okay, I get everything up to this point

  18. Nnesha
    • one year ago
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    to write an equation we have to find 1) distance between the point `on parabola ` to the focus 2nd) distance between the point " on parabola" to the directrix

  19. anonymous
    • one year ago
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    Okay is there a formula for this?

  20. Nnesha
    • one year ago
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    |dw:1444449758151:dw| blue line distance ( between point and directrix ) red one distance between (point and focus ) and the equation i posted above \[\huge\rm \sqrt{(x-a)^2 +(y-b)^2 }= \left| y-c \right|\]

  21. Nnesha
    • one year ago
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    (a,b) is the focus point c= directrix

  22. Nnesha
    • one year ago
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    now plugin the numbers (a,b) = (1,6) c= -2

  23. anonymous
    • one year ago
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    so sqrt of (x - 1)^2 + (y - 6)^2 = (y + 2)^2 right?

  24. Nnesha
    • one year ago
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    hmm \[\huge\rm \sqrt{(x-1)^2 +(y-6)^2 }= \left| y+2 \right|\] take square both sides please let me know if you don't get anything i'm sleepy e,e

  25. anonymous
    • one year ago
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    Ill let you know, thanks for helping! And what do you mean by "take square both sides" ?

  26. Nnesha
    • one year ago
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    when we take square both sides tit would be \[\huge\rm (x-1)^2 +(x-6)^2 = (y+2)^2\]

  27. anonymous
    • one year ago
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    ohh okay, right. to get rid of the sqrt and the absolutes

  28. Nnesha
    • one year ago
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    yes right

  29. anonymous
    • one year ago
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    and then what??

  30. Nnesha
    • one year ago
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    and then foil (x-6)^2 is same as (x-6)(x-6) same with (y+2)^2 bec as dan mentioned we need an equation y= form

  31. Nnesha
    • one year ago
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    ugh typo it's \[\huge\rm (x-1)^2 +(y-6)^2 = (y+2)^2\]

  32. anonymous
    • one year ago
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    do i foil x-1 ^2 too?

  33. Nnesha
    • one year ago
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    sorry sorry :( (y-6)(y-6)

  34. Nnesha
    • one year ago
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    well i'll saw leave it as (x-h)^2 we can write an equation in vertex form \[y=a(x-h)^2+k\]

  35. anonymous
    • one year ago
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    okay so (y^2 - 12y + 36) + (x-1)^2 = y^2 +4y + 4

  36. Nnesha
    • one year ago
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    right now combine like terms |dw:1444450916236:dw|

  37. Nnesha
    • one year ago
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    |dw:1444450996389:dw|

  38. Nnesha
    • one year ago
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    sorry about the cool writing ,-,

  39. anonymous
    • one year ago
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    wait, that last thing threw me off

  40. Nnesha
    • one year ago
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    ohh okay which one please ask :=)

  41. Nnesha
    • one year ago
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    i just moved all terms to the right side except (x-1)^2

  42. anonymous
    • one year ago
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    oh never mind, i get it! Sorry

  43. Nnesha
    • one year ago
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    okay thats fine

  44. anonymous
    • one year ago
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    Okay and now what do i do?

  45. Nnesha
    • one year ago
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    let me think ... :( i'm lost :( wait a sec plz

  46. anonymous
    • one year ago
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    Okkay thats fine!

  47. anonymous
    • one year ago
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    I just submitted the exam because i have to go to bed. I got the answer wrong too :/ but thanks for all your help!

  48. Nnesha
    • one year ago
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    alright i'll going to expand (x-1)^2 \[\large\rm x^2-2x+1 +y^2-12y+36=y^2+4y+4\] |dw:1444451678408:dw|

  49. Nnesha
    • one year ago
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    that's 16 i'm so sorry

  50. Nnesha
    • one year ago
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    i shouldn't start helping u so my fault i apologize i'll look at it tomorrow will try to explain and i wil lfocus on this

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