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anonymous

  • one year ago

help me

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  1. anonymous
    • one year ago
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    @LynFran

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

  3. anonymous
    • one year ago
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    ??

  4. Nnesha
    • one year ago
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    is it horizontal ellipse or vertical ellipse?

  5. anonymous
    • one year ago
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    idk lol

  6. LynFran
    • one year ago
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    c^2=a^2-b^2

  7. Nnesha
    • one year ago
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    if bigger number under the x coordinate then horizontal and if it's under the y coordiante then vertical

  8. anonymous
    • one year ago
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    so its horizontal

  9. Nnesha
    • one year ago
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    yes right so thats mean you should add c into x coordinate of the center \[\huge\rm (h \pm c,k)\] now use that equation lynn gave u to find c value

  10. Nnesha
    • one year ago
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    (h,k) is the center

  11. anonymous
    • one year ago
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    ok

  12. Nnesha
    • one year ago
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    btw standard form of ellipse is \[\huge\rm \frac{ (x-h)^2 }{ a^2 } + \frac{ (y-k)^2 }{ b^2 }=1\] a^2= ?? b^2= ?? in ur question

  13. anonymous
    • one year ago
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    yeah how do i found this out

  14. Nnesha
    • one year ago
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    they already in the question a^2 is under the x b^2=under the y

  15. anonymous
    • one year ago
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    ok sooooo

  16. Nnesha
    • one year ago
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    substitute a^2 and b^2 value into the equation lynn gave you c^2=a^2-b^2 then solve for c

  17. anonymous
    • one year ago
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    so hard im just now learing this lol

  18. anonymous
    • one year ago
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    ohh

  19. Nnesha
    • one year ago
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    thats fine \[\huge\rm \frac{ (x-h)^2 }{ \color{ReD}{a^2 }} + \frac{ (y-k)^2 }{\color{ReD}{ b^2 }}=1\] \[\huge\rm \frac{ x^2 }{\color{Red}{ 36} }+\frac{ y^2 }{\color{ReD}{ 11} }=1\] bg number would be `a^2` always!(in ellipse equation)

  20. Nnesha
    • one year ago
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    so a^2 = ?

  21. anonymous
    • one year ago
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    1

  22. Nnesha
    • one year ago
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    no a^2 is under the x^2 coordinate

  23. Nnesha
    • one year ago
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    http://prntscr.com/89k6q9

  24. anonymous
    • one year ago
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    so36

  25. Nnesha
    • one year ago
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    yes right now what is b^2=?

  26. anonymous
    • one year ago
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    11

  27. Nnesha
    • one year ago
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    perfect! \[\huge\rm c^2=a^2-b^2\] substitute a^2 and b^2 for their values solve for c

  28. anonymous
    • one year ago
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    c^2=-1175

  29. anonymous
    • one year ago
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    am i right

  30. Nnesha
    • one year ago
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    ayooo how :o

  31. Nnesha
    • one year ago
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    a^2=36 b^2=11 it's already squared you don't need to square them again

  32. anonymous
    • one year ago
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    sooo

  33. Nnesha
    • one year ago
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    so.. substitute a^2 for 36 and b^2 for 11

  34. anonymous
    • one year ago
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    25

  35. anonymous
    • one year ago
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    so c^2=25

  36. Nnesha
    • one year ago
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    yes right now take square root both sides to cancel out the square

  37. anonymous
    • one year ago
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    so c=5

  38. Nnesha
    • one year ago
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    yes right now what's the center point ?

  39. anonymous
    • one year ago
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    idk

  40. Nnesha
    • one year ago
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    \[\huge\rm \frac{ (x-\color{reD}{h})^2 }{ a^2 } + \frac{ (y-\color{ReD}{k})^2 }{ b^2 }=1\] are there any h and k value at the numerator ?

  41. anonymous
    • one year ago
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    ??

  42. Nnesha
    • one year ago
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    look at the original equation is there any number with x ?

  43. anonymous
    • one year ago
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    yeah its sqared

  44. Nnesha
    • one year ago
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    not square (x-3)^2 then h would be 3 for this equation center point is (0,0)

  45. Nnesha
    • one year ago
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    so formula for foci is \[\huge\rm (h \pm c,k)\] where (h,k) is the center (0,0) c=5 substitute variables for values

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