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Math Help - Finding the equation of an ellipse

  1. #1
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    Finding the equation of an ellipse

    Hello,
    I have an assigned question which asks:

    In the x-y plane let L be the line x = 1, let the point O be the origin,
    and set L = 1

    Find an equation for the set of points P = (x, y) such that the
    ratio of the distances |OP|/|PM|= 1/2 where M is the point on
    the line L that is closest to P. Note that from the class lecture on
    the Kepler problem, you know that this curve is an ellipse. Since
    one of the foci of the ellipse is at the origin, the equation may
    not look like what you are used to.

    Our lecture did not cover how to find the equation of the ellipse however if briefly described it and I have been looking in my text book for help but with no luck. If anyone could give me any advice and help me out I would really appreciate it.
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  2. #2
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    Re: Finding the equation of an ellipse

    Hello, lytwynk!

    \text{In the }xy\text{-plane, let }O\text{ be the origin and let }L\text{ be the line }x = 1.

    \text{Find an equation for the set of points }P(x, y)
    \text{such that the ratio of the distances: }\:\frac{|OP|}{|PM|} \,=\, \frac{1}{2}
    \text{where }M\text{ is the point on line }L\text{ that is closest to }P.

    \text{Note that from the class lecture on the Kepler problem, you know that this curve is an ellipse.}
    \text{Since one of the foci of the ellipse is at the origin, the equation may not look like what you are used to.}

    The diagram looks like this:

    Code:
            |     P     :M
            |(x,y)* - - *(1.y)
            |    /      :
            |   /       :
            |  /        :
            | /         :
            |/          :
        - - * - - - - - + - - -
            |O          1
            |
    OP \:=\:\sqrt{x^2+y^2}

    PM \:=\:1-x

    \text{W\!e have: }\:\sqrt{x^2+y^2} \:=\:\tfrac{1}{2}(1-x) \quad\Rightarrow\quad 2\sqrt{x^2+y^2} \:=\:1 - x

    \text{Square both sides: }\:4(x^2+y^2) \:=\:(1-x)^2 \quad\Rightarro\quad 4x^2 + 4y^2 \:=\:1-2x +x^2

    . . 3x^2 + 2x + 4y^2 \:=\:1 \qiad\Rightarrow\quad 3\left(x^2 + \tfrac{2}{3}x\right) \,+\,4y^2 \:=\:1

    . . 3\left(x^2 + \tfrac{2}{3}x + \tfrac{1}{9}\right) + 4y^2 \:=\:1 + \tfrac{1}{3} \quad\Rightarrow\quad 3\left(x + \tfrac{1}{3}\right)^2 + 4y^2 \:=\:\tfrac{4}{3}

    \text{Divide by }\tfrac{4}{3}:\;\;\frac{3(x+\frac{1}{3})^2}{\frac{  4}{3}} +  \frac{4y^2}{\frac{4}{3}} \:=\:\frac{\frac{4}{3}}{\frac{4}{3}}

    \text{And we have: }\:\frac{(x+\frac{1}{3})^2}{\frac{4}{9}} + \frac{y^3}{\frac{1}{3}} \:=\:1


    . . . \begin{Bmatrix}\text{Center: }\:(\text{-}\frac{1}{3},\,0) \\ \\[-3mm]  \text{Semimajor axis: }\: a \,=\,\frac{2}{3} \\ \\[-3mm]  \text{Semiminor axis: }\:b \,=\,\frac{\sqrt{3}}{3}\end{Bmatrix}
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  3. #3
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    Re: Finding the equation of an ellipse

    Thanks so much, this was a huge help.
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