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Math Help - Parametric Equation of Parabola

  1. #1
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    Parametric Equation of Parabola

    Hello, just a question.

    Find the cartesian equation of the curves whose parametric equations are:

    a) x = 2cos(theta), y = 2sin(theta), 0 ≤ theta ≤ pi

    the answer is: y = sqrt( 4 - x^2), -2 ≤ x ≤ 2


    I understand how to do it up to x^2 + y^2 = 4, and I can arrive at the answer, but I do not know WHY the solution is y = sqrt( 4 - x^2), -2 ≤ x ≤ 2. Something to do with the domain, but I dont understand.

    Thanks.
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  2. #2
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    Quote Originally Posted by noobonastick View Post
    I understand how to do it up to x^2 + y^2 = 4, and I can arrive at the answer, but I do not know WHY the solution is y = sqrt( 4 - x^2), -2 ≤ x ≤ 2. Something to do with the domain, but I dont understand.
    I am not too sure which part you are having trouble with, so I will go through my own solution.

    x=2\cos\theta,\;y=2\sin\theta\quad0\leq\theta\leq\  pi

    We have

    x^2+y^2=(2\cos\theta)^2+(2\sin\theta)^2=4\cos^2\th  eta+4\sin^2\theta (just substituting for x and y)

    \Rightarrow x^2+y^2=4. (Pythagorean identity)

    But note that y must be nonnegative (since 2\sin\theta is nonnegative when 0\leq\theta\leq\pi). Similarly, -2\leq x\leq2. So, we can solve for y, but we must throw out the negative root as it is not part of the original curve. So we get

    y^2=4-x^2

    \Rightarrow y=\pm\sqrt{4-x^2}

    \Rightarrow y=\sqrt{4-x^2} (because y\geq0)

    and our final answer is

    y=\sqrt{4-x^2},\quad-2\leq x\leq2.
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  3. #3
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    By the way, that's not a parabola, its a circle.

    After you got x^2+ y^2= 4, you need to recognise that y= sin(\theta) is non-negative for 0\le \theta\le \pi and do you can solve for y taking the positive root.
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