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Math Help - Definine a Formula as a Parabola

  1. #1
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    Definine a Formula as a Parabola

    Hello,

    How would I go about showing that the equation (x-a)^2 = x^2 + y^2 defines a parabola? If I put the equation in terms of y, I get y = sqrt(a^2 - 2xa), which is not a parabola. I've found the derivative to be y' = -a/y, but I don't think that proves anything. If anyone has any ideas, I'd really appreciate them! Thanks.

    PS 'a' is a constant.
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  2. #2
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    Quote Originally Posted by Dudealadude View Post
    Hello,

    How would I go about showing that the equation (x-a)^2 = x^2 + y^2 defines a parabola? If I put the equation in terms of y, I get y = sqrt(a^2 - 2xa), which is not a parabola. I've found the derivative to be y' = -a/y, but I don't think that proves anything. If anyone has any ideas, I'd really appreciate them! Thanks.

    PS 'a' is a constant.
    x^2-2ax+a^2=x^2+y^2\Rightarrow\ 2ax=-y^2+a^2\Rightarrow\ x=\displaystyle-\frac{1}{2a}y^2+\frac{a}{2}

    which is a parabola.
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  3. #3
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    Quote Originally Posted by Dudealadude View Post
    Hello,

    How would I go about showing that the equation (x-a)^2 = x^2 + y^2 defines a parabola? If I put the equation in terms of y, I get y = sqrt(a^2 - 2xa), which is not a parabola. I've found the derivative to be y' = -a/y, but I don't think that proves anything. If anyone has any ideas, I'd really appreciate them! Thanks.

    PS 'a' is a constant.
    x^2 - 2ax + a^2 = x^2 + y^2

    2ax = a^2-y^2

    x = \frac{a}{2} - \frac{1}{2a} y^2

    parabola, line of symmetry is the x-axis ... opens left from vertex \left(\frac{a}{2} , 0 \right)

    graph shows parabola for a = 2
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  4. #4
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    y = sqrt(a^2 - 2xa)
    Actually, the equation y = \hspace{2mm} $\pm$ \sqrt{a^2 - 2xa} is a parabola. You probably don't recognize it as such because it opens to the left, rather than up or down. You can show that it's a parabola by comparing it to the conic section equation:

    Definine a Formula as a Parabola-47bff2d8bfe4b12d6a4cd286bc99f11e.png
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