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Conversion of polar equation to rectangular equation

  1. May 31, 2009 #1
    r= (15)/(3-2cos(theta))


    I'm lost!!
    Please Help!!
    :confused:
     
  2. jcsd
  3. May 31, 2009 #2
    in polar coordinates r=(x^2+y^2)^1/2 and also x=rcosQ ,y=rsinQ so putting these in above equation and cosQ=x/r we get( x^2+y^2)^1/2=15/(3-2*x/( x^2+y^2)^1/2) now this can be solved
     
  4. May 31, 2009 #3
    In case Vandanak's brilliant solution is hard to read, I've formatted his statements in TeX

    Given:
    [tex] r = \frac{15}{3-2cos(\theta)} [/tex]

    [tex] r = \sqrt{x^2+y^2} [/tex]

    [tex] x = rcos(\theta) [/tex]

    [tex] y = rsin(\theta) [/tex]

    substituting
    [tex]\sqrt{x^2+y^2} = \frac{15}{3-2(\frac{x}{\sqrt{x^2+y^2}})} [/tex]
     
  5. Jun 1, 2009 #4

    arildno

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    If I did this correctly, you should end up with:
    [tex]y=\pm\frac{\sqrt{5(15-x)(x+3)}}{3}[/tex]

    Thus, the solution is a closed curve.
     
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