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Math Help - Help with Parametric Equation

  1. #1
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    Help with Parametric Equation

    I'm an Algebra 2 student and we are learning parametric equations. I am having difficulty solving this one, although I'm good on most of the others.

    Here it is:

    y = 2t
    x = t^2 + 2t

    The trouble I'm having is coming up with a "y=" equation, because the y term becomes quadratic. What I have gotten down to is: y^2 = 4x - 4y. Where do I go from here? Thanks...
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  2. #2
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    An easy way of solving this is solving for t on the x=t^2+2t equation and putting into the other equation y=2t
    you have to solve it by completing the square on x=t^2+2t
    so after doing that, you get x+1=(t+1)^2
    then square root both sides +√(x+1)=t+1 and -√(x+1)=t+1
    then -1 to both sides t=-1+√(x+1) and t=-1-√(x+1)
    substitute both of those into the equation y=2t
    and you get y=2(-1+√(x+1)) and y=2(-1-√(x+1))
    simplified, you get y=-2+2√(x+1) and y=-2-2√(x+1)

    you get two equations because when you graph the parametric equation, you get a sideways parabola, therefore, it is not a function, and has to be split into two different functions so it is possible to be graphed.
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