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Polar Forms

  1. Aug 26, 2005 #1
    I am working a problem:
    Find the polar form of the equation 6x^2 + 3xy + 6y^2 = -3
    I have run this 4 separate times and have come up with 4 different answers.

    PLEASE HELP!
    r^2= 6 over -6 - 3sin(theta)cos(theta)

    r^2=-3 over 6 - 3sin(theta)cos(theta)

    r^2=9 over 6 + 3sin(theta)cos(theta)

    r^2=-3 over 6 + 3sin(theta)cos(theta)
    :cry:
     
  2. jcsd
  3. Aug 26, 2005 #2

    lurflurf

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    Homework Helper

    Don't make things hard.
    recall (i an using t instead of theta)
    x^2+y^2=r^2
    2xy=r^2sin(2t)
    we have
    6x^2 + 3xy + 6y^2 = -3
    get nice numbers by multiplying by 2/3
    4x^2 + 2xy + 4y^2 = -2
    group by know facts
    4(x^2+y^2)+2xy=-2
    use known fact
    4r^2+r^2sin(2t)=-2
    factor r^2
    r^2(4+sin(2t))=-2
    divide
    r^2=-2/(4+sin(2t))
    remember different equation can mean the same thing
    I think some of your answers may be correct, but I have not checked them in detail.
    r^2= 6 over -6 - 3sin(theta)cos(theta)
    r^2=-3 over 6 + 3sin(theta)cos(theta)
    are equivalent to my answer
    keep in mind we have r^2<0 hence r is and imaginary number
    the original equation can be written
    3x^2+(x+y)^2 + 3y^2 = -2
    clearly this has no solution when x and y are restricted to real numbers.
     
  4. Aug 26, 2005 #3
    Rearrange your equation into the form [itex]6(x^2+y^2) + 3xy + 3 = 0 [/itex] , now do the following substitutions:

    x=r cosQ
    y=r sinQ

    and

    [itex]r^2 = x^2 + y^2[/itex]

    BJ
     
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