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Math Help - Polar equation

  1. #1
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    Polar equation

    I need help with this one:

    Find an equation in x and y for the conic section with polar equation r=
    1
    ___
    1+cos[theta]

    possible answers:
    y[squared]=1-2x
    x[squared]-x+y[squared]=1
    x[squared]+y[squared]=1
    x[squared]+x-y=1
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by mike1 View Post
    I need help with this one:

    Find an equation in x and y for the conic section with polar equation r=
    1
    ___
    1+cos[theta]

    possible answers:
    y[squared]=1-2x
    x[squared]-x+y[squared]=1
    x[squared]+y[squared]=1
    x[squared]+x-y=1
    Well,
    x = r*cos(t)
    y = r*sin(t)

    Thus x^2 + y^2 = r^2*cos^2(t) + r^2*sin^2(t) = r^2(cos^2(t) + sin^2(t)) = r^2

    So r = sqrt{x^2 + y^2}

    Thus your equation is
    sqrt{x^2 + y^2} = 1

    Squaring both sides gives,
    x^2 + y^2 = 1, which is your third option.

    -Dan
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  3. #3
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    Hello, Mike!

    Find an equation in x and y for the conic section with polar equation:
    . . . . . . . . . 1
    . . r .= . -----------
    . . . . . . 1 + cos θ

    Possible answers:

    . . y .= .1 - 2x . . x - x + y .= .1 . . x + y .= .1 . . x + x - y .= .1

    We have: .r(1 + cosθ) .= .1

    . . . . . . . . .r + rcosθ .= .1

    . . . . . . . . . . . . . . .r .=] .1 - rcosθ
    . . . . . . . . . . ______
    Convert: . . .√x + y .= .1 - x

    Square: . . . . .x + y .= .1 - 2x + x

    Therefore: . . . . . . y .= .1 - 2x



    (Dan, you got a unit circle. .That equation would be: r = 1)
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  4. #4
    Forum Admin topsquark's Avatar
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    I didn't see the 1 + cos(t) part.

    -Dan
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