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Thread: Find centre of ellipse (Polar)

  1. #1
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    Find centre of ellipse (Polar)

    Show that the centre C of the ellipse $\displaystyle r = \frac{ep}{1 + ecos(\theta)} $ has the coordinates $\displaystyle \lgroup \frac{e^2p}{1 - e^2}, -\pi \rgroup $

    Where e is the eccentricity and p the distance from focus to directrix.

    Here is my attempt

    The vertices of the major axis occur at 0 and $\displaystyle \pi $

    The vertices of the minor axis occur at $\displaystyle \frac{\pi}{2} $ and $\displaystyle \frac{-\pi}{2} $

    First coordinate by midpoint rule $\displaystyle \frac{\frac{ep}{1 - e} + \frac{ep}{1 + e}}{2} = \frac{\frac{2ep}{1 - e^2}}{2} = \frac{ep}{1 - e^2} $

    By similar reasoning, second coordinate by midpoint rule yields ep.

    So I get $\displaystyle C\lgroup \frac{ep}{1 - e^2}, ep \rgroup $

    Where have I gone wrong?
    Last edited by Altmer; Feb 1st 2019 at 11:09 AM.
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  2. #2
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    Re: Find centre of ellipse (Polar)

    Quote Originally Posted by Altmer View Post
    Show that the centre C of the ellipse $\displaystyle r = \frac{ep}{1 + ecos(\theta)} $ has the coordinates $\displaystyle \lgroup \frac{e^2p}{1 - e^2}, -\pi \rgroup $

    Where e is the eccentricity and p the distance from focus to directrix.

    Here is my attempt

    The vertices of the major axis occur at 0 and $\displaystyle \pi $

    The vertices of the minor axis occur at $\displaystyle \frac{\pi}{2} $ and $\displaystyle \frac{-\pi}{2} $

    First coordinate by midpoint rule $\displaystyle \frac{\frac{ep}{1 - e} + \frac{ep}{1 + e}}{2} = \frac{\frac{2ep}{1 - e^2}}{2} = \frac{ep}{1 - e^2} $

    By similar reasoning, second coordinate by midpoint rule yields ep.

    So I get $\displaystyle C\lgroup \frac{ep}{1 - e^2}, ep \rgroup $

    Where have I gone wrong?
    You are attempting to use the midpoint formula (for x-y coordinates) for polar coordinates
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  3. #3
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    Re: Find centre of ellipse (Polar)

    Quote Originally Posted by Altmer View Post
    Show that the centre C of the ellipse $\displaystyle r = \frac{ep}{1 + ecos(\theta)} $ has the coordinates $\displaystyle \lgroup \frac{e^2p}{1 - e^2}, -\pi \rgroup $

    Where e is the eccentricity and p the distance from focus to directrix.

    Here is my attempt

    The vertices of the major axis occur at 0 and $\displaystyle \pi $

    The vertices of the minor axis occur at $\displaystyle \frac{\pi}{2} $ and $\displaystyle \frac{-\pi}{2} $

    First coordinate by midpoint rule $\displaystyle \frac{\frac{ep}{1 - e} + \frac{ep}{1 + e}}{2} = \frac{\frac{2ep}{1 - e^2}}{2} = \frac{ep}{1 - e^2} $

    By similar reasoning, second coordinate by midpoint rule yields ep.

    So I get $\displaystyle C\lgroup \frac{ep}{1 - e^2}, ep \rgroup $

    Where have I gone wrong?
    subtract rather than add since the second vertex is on the negative x-axis
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