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Math Help - Polar Coordinates

  1. #1
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    Polar Coordinates

    Hi. I need to figure out how to convert the polar equation
    r = 2cos \theta - sin \theta
    to cartesian coordinates and then graph the resulting equation.

    I know it's an ellipse and I know x = rcos \theta, y = rsin \theta and r^2 = x^2 + y^2, but I can't quite figure out how to make this equation into the standard form for an ellipse. The closest I've come is \frac{x^2-2x}{2} + \frac{y^2 + y}{2} = 0, which doesn't make any sense. Could someone help me? (Also, I wasn't sure whether this belonged in the calculus or geometry forum, but it's for my calc class so I posted it here.)
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  2. #2
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    Anchorage, AK
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    First, multiply both sides of your polar equation by r:
    r^2=2r\cos\theta-r\sin\theta
    Now, use x=r\cos\theta, y=r\sin\theta, and r^2=x^2+y^2, like you said, to get:
    x^2+y^2=2x-y
    Rearranging:
    x^2-2x+y^2+y=0
    Completing the square on both the x and y terms:
    x^2-2x+1+y^2+y+\frac{1}{4}=\frac{5}{4}
    (x-1)^2+\left(y+\frac{1}{2}\right)^2=\frac{5}{4}
    You should be able to recognise what sort of conic section this is, and be able to graph it.

    --Kevin C.
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  3. #3
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    Quote Originally Posted by TwistedOne151 View Post
    First, multiply both sides of your polar equation by r:
    r^2=2r\cos\theta-r\sin\theta
    Now, use x=r\cos\theta, y=r\sin\theta, and r^2=x^2+y^2, like you said, to get:
    x^2+y^2=2x-y
    Rearranging:
    x^2-2x+y^2+y=0
    Completing the square on both the x and y terms:
    x^2-2x+1+y^2+y+\frac{1}{4}=\frac{5}{4}
    (x-1)^2+\left(y+\frac{1}{2}\right)^2=\frac{5}{4}
    You should be able to recognise what sort of conic section this is, and be able to graph it.

    --Kevin C.
    Thank you Kevin. I always forget to complete the square.
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