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anonymous

  • one year ago

help me plz

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  1. anonymous
    • one year ago
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  2. anonymous
    • one year ago
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    @Directrix

  3. anonymous
    • one year ago
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    brb in 20

  4. anonymous
    • one year ago
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    im back

  5. Nnesha
    • one year ago
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    first of all you should know is it horizontal or vertical a always greater than b `a>b` if a is under the x variable then it would be `horizontal equation` \[\rm \frac{ (x-h)^2 }{\color{ReD}{ a^2} }+ \frac{ (y-k)^2 }{ b^2 }=1\] if a is under the y variable then it would be `vertical equation` \[\rm \frac{ (x-h)^2 }{ b^2 }+ \frac{ (y-k)^2 }{\color{reD}{ a^2} }=1\]

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

  7. anonymous
    • one year ago
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    i think its vertical

  8. Nnesha
    • one year ago
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    foci for horizontal \[\large\rm (h, k \pm c)\] and for vertical equation \[\large\rm (h \pm c , k)\] use the equation to find c \[\huge\rm c^2 =a^2-b^2\]

  9. anonymous
    • one year ago
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    ok so which one do i use

  10. Nnesha
    • one year ago
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    well a is bigger number which is under the x variable so is it vertical or horizontal ?

  11. Nnesha
    • one year ago
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    read my first comment :=)

  12. anonymous
    • one year ago
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    so it is horizontal right

  13. Nnesha
    • one year ago
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    yes right \[\huge\rm \frac{ x^2 }{ \color{ReD}{36} } +\frac{ y^2 }{ 11 }=1\] now use the equation posted above to find c

  14. Nnesha
    • one year ago
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    c^2=a^2-b^2 this one

  15. anonymous
    • one year ago
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    wait what

  16. Nnesha
    • one year ago
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    foci for horizontal \[\large\rm (h, k \pm c)\] and for vertical equation \[\large\rm (h \pm c , k)\] those two are foci pair we need so use this equation to find c \[\huge\rm c^2 =a^2-b^2\]

  17. Nnesha
    • one year ago
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    a^2 and b^2 are in the given equation

  18. anonymous
    • one year ago
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    wow ok so umm can you start me off plz

  19. Nnesha
    • one year ago
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    `horizontal equation` for ellipse \[\rm \frac{ (x-h)^2 }{\color{ReD}{ a^2} }+ \frac{ (y-k)^2 }{ b^2 }=1\] the number at the denominator are a^2 and b^2 and like i said bigger number equal to a^2

  20. Nnesha
    • one year ago
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    what two numbers are replaced with a^2 and b^2 in this equation \[\huge\rm \frac{ x^2 }{ \color{ReD}{36} } +\frac{ y^2 }{ 11 }=1\] ??

  21. anonymous
    • one year ago
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    brb i gtg cook dinner

  22. anonymous
    • one year ago
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    36 and 11

  23. Nnesha
    • one year ago
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    right which one is a^2 and which one is b^2 ?

  24. anonymous
    • one year ago
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    a^2 is 36 b^2 is 11

  25. anonymous
    • one year ago
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    @IrishBoy123

  26. anonymous
    • one year ago
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    { (x-h)^2 }{36} }+/{ (y-k)^2 }{11}=1 is this the correct formula

  27. IrishBoy123
    • one year ago
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    this ellipse is centred on the origin yep?

  28. anonymous
    • one year ago
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    yep

  29. IrishBoy123
    • one year ago
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    |dw:1444085734817:dw|

  30. anonymous
    • one year ago
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    yep

  31. IrishBoy123
    • one year ago
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    Para-phrasing Wiki: The distance from the origin O to either focus is f where: \[f = \sqrt{a^2-b^2}.\]

  32. anonymous
    • one year ago
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    ok she gave me a diffrent formula soo what do i do im so confused now

  33. IrishBoy123
    • one year ago
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    here, what is a and what is b? general equation for ellipse centred on Origin is: \(\large \frac{{x}^2}{{a}^2} + \frac{{y}^2}{{b}^2} = 1\)

  34. IrishBoy123
    • one year ago
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    a = ?? b = ??

  35. IrishBoy123
    • one year ago
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    then |dw:1444086259207:dw|

  36. anonymous
    • one year ago
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    (x-h)^2 }{36} }+/{ (y-k)^2 }{11}= ok so i got this

  37. IrishBoy123
    • one year ago
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    so that means you have finished it?!

  38. anonymous
    • one year ago
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    no i have not thats the first equation

  39. anonymous
    • one year ago
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    (y-k)^2 {11} this is my answer

  40. IrishBoy123
    • one year ago
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    OK for an equation centred on O, we have general equation: \[\large \frac{{x}^2}{{a}^2} + \frac{{y}^2}{{b}^2} = 1\] The focii are at a distance f from O, where \[f = \sqrt{a^2-b^2}\] |dw:1444086808815:dw|

  41. IrishBoy123
    • one year ago
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    soz, we crossed posts. i understood the question to be finding the focii of the ellipse

  42. anonymous
    • one year ago
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    yes idk what im doing im so cunfused here im sorrry

  43. IrishBoy123
    • one year ago
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    no worries you have here: \(\large\rm \frac{ x^2 }{ \color{ReD}{6^2} } +\frac{ y^2 }{(\sqrt{\color{red} 1\color{red}1 })^2}=1\) and generally speaking: \(\large \frac{{x}^2}{{a}^2} + \frac{{y}^2}{{b}^2} = 1\) and: \(\large f = \sqrt{a^2-b^2}\)

  44. IrishBoy123
    • one year ago
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    so f =????

  45. anonymous
    • one year ago
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    omg i have no clue on how to do this like wth

  46. anonymous
    • one year ago
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    does f equal -4

  47. IrishBoy123
    • one year ago
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    \(a = 6, \; b = \sqrt{11}\) \(f = \sqrt{ 6^2 - (\sqrt{11})^2}\) \(f = 5\)

  48. anonymous
    • one year ago
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    atleast i was close haha

  49. IrishBoy123
    • one year ago
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    finished ?? :p

  50. anonymous
    • one year ago
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    i think so

  51. IrishBoy123
    • one year ago
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    cool!! well done

  52. anonymous
    • one year ago
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    thnx for your help

  53. IrishBoy123
    • one year ago
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    mp

  54. Nnesha
    • one year ago
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    \(\color{blue}{\text{Originally Posted by}}\) @Tazmaniadevil ok she gave me a diffrent formula soo what do i do im so confused now \(\color{blue}{\text{End of Quote}}\) the formula is same difference is f and c variables f= foci \[\huge\rm f = \sqrt{ a^2 -b^2}\] is same as \[\huge\rm c^2= a^2-b^2 ~~~~~c=\sqrt{a^2-b^2}\]

  55. Nnesha
    • one year ago
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    a^2 and b^2 are already given so would be easy you write it as c= sqrt{36-11} :=)

  56. anonymous
    • one year ago
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    OK

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