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Conversion from Polar to Cartesian (ellipse)

  1. Nov 4, 2007 #1
    1. The problem statement, all variables and given/known data
    Convert the conic section to standard form. r=[tex]\frac{1}{8-4*sin(\theta}[/tex]

    2. Relevant equations
    x=rcos(\theta)
    y=rsinx(\theta)

    3. The attempt at a solution

    r=[tex]\frac{1}{8-4*sin(\theta}[/tex]

    [tex]r^2=\frac{1}{64-64*sin(\theta)+16sin^2(\theta)} [/tex]

    r^2= x^2 + y^2

    I can see the y=r*sin(\theta) but not the x!
     
  2. jcsd
  3. Nov 5, 2007 #2
    so what does r= according to the definitions and what does sin(theta)= according to the definitions? What can you do with that knowledge?
     
    Last edited: Nov 5, 2007
  4. Nov 5, 2007 #3
    I would multiply both sides by (8-4sin(theta)) and then you can replace rsin(theta) with y. And then you can use your identity r^2 = x^2 + y^2.
     
  5. Nov 5, 2007 #4
    I came up with [tex]r= \sqrt{x^2 + y^2}[/tex] and [tex]sin(\theta)=y/r[/tex]

    [tex]\sqrt{x^2 + y^2} = \frac{1}{8-4sin(\theta)}[/tex]

    [tex]\sqrt{x^2 + y^2} = \frac{1}{8-(\frac{4y}{r})}[/tex]

    [tex]x^2 + y^2 = 64 - \frac{64y}{r} + \frac{16y^2}{r^2}[/tex]

    pretty sure I chose the wrong order of events there...



    But when I tried the other suggestion I got pretty close to an answer:

    [tex]8r - 4rsin(\theta) = 1[/tex]

    [tex]8r - 4y = 1[/tex]

    [tex]8r = 1 + 4y[/tex]

    [tex]r = \frac{1+4y}{8}[/tex]

    [tex]r^2 = \frac{1+8y+16y^2}{64}[/tex]

    [tex]x^2 + y^2 = \frac{1+8y+16y^2}{64}[/tex]

    [tex]64x^2 + 64y^2 = 1+8y+16y^2[/tex]

    [tex]64x^2 + 48y^2 - 8y = 1[/tex]

    and this is where I get stuck because 48y^2 - 8y +/- ___
     
  6. Nov 5, 2007 #5
    r= r/(8r-4y)

    8r-4y=1
     
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