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Math Help - Eccentricity, trouble w/ the proof

  1. #1
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    Eccentricity, trouble w/ the proof

    I'm trying to understand the proof for eccentricity; I can get about halfway through it. The problem I'm having is summarized below

    If e<1, you get the equation of an ellipse of the form

    \frac{(x-h)^2}{a^2}+\frac{y^2}{b^2}=1

    where

    Eq. 4:
    h= -\frac{e^2d}{1-e^2}
    a^2= -\frac{e^2d^2}{(1-e^2)^2}
    b^2= -\frac{e^2d^2}{1-e^2}

    The foci of an ellipse are at a distance c from the center, where

    Eq 5:
     c^2=a^2-b^2= \frac{e^4d^2}{(1-e^2)^2}

    This shows that

    c= \frac{e^2d}{1-e^2}=-h

    It follows from equations 4 and 5 that the eccentricity is given by

    e=\frac{c}{a}

    If someone could help explain how e=\frac{c}{a} is derived, or really what it means I'd be so grateful. I'm using Calculus Early Transcendentals 6th Ed. by Stewart (Ch. 10, Sect. 6).
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  2. #2
    MHF Contributor Siron's Avatar
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    Re: Eccentricity, trouble w/ the proof

    The conclusion is that the eccentricity of an ellips is: e=\frac{c}{a}, now:
    If you write: e^2=\frac{c^2}{a^2}
    Calculating with this:
    \frac{c^2}{a^2}=\frac{a^2-b^2}{a^2}=1-\frac{b^2}{a^2}
    Substitution of b^2 and a^2 out of eq 4 and you get:
    1-\frac{b^2}{a^2}=1-\left[\frac{\frac{-e^2d^2}{1-e^2}}{\frac{-e^2d^2}{(1-e^2)^2}}}\right]
    =1-(1-e^2)=e^2
    Now take the square root of e^2=e.
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