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Math Help - Why doesn't my ellipse inversion work gosh darn it?

  1. #1
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    Why doesn't my ellipse inversion work gosh darn it?

    Hi,

    Please see the attached images.

    I am trying to invert an ellipse about one of its foci. I am not sure if I'm doing it right. My resulting inversion (image on the right) should look like the textbook image on the left, but seems to differ slightly.

    In both images the limacon intercepts the ellipse at the point where the circle of inversion intercepts the ellipse. But in my inversion this point also happens to be in line with the center of the ellipse (as indicated by the dotted line). This is different from the textbook case.

    Am I doing something wrong, or is this difference just owing to the different parameters of the two ellipses?

    The ellipse I want to invert is: \frac{(x+b)^2}{a^2}+\frac{y^2}{b^2}=1

    b, the semi-minor axis = \frac{\sqrt{2}}{2}

    a, the semi-major axis = 1

    The focus about which I want to invert is located at the origin (0,0).

    To invert all of the points on the ellipse about the focus I am
    using this equation:

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

    (I'm not 100% sure if this is the correct equation for inversion)
    Attached Thumbnails Attached Thumbnails Why doesn't my ellipse inversion work gosh darn it?-ellipse-inversion-xahlee-website.bmp   Why doesn't my ellipse inversion work gosh darn it?-eillipse-inversion-example.jpg  
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  2. #2
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    The fact that the points of intersection are in line with the centre of the ellipse is just a coincidence caused by the choice of parameters. The unit circle centred at the origin goes through the points (-1/\sqrt2,\pm1/\sqrt2), and these happen to be the ends of the minor axis of the ellipse.

    But the equation of the limašon is wrong, and the graph of it looks squashed up. It should extend out much further to the right of the origin. The equation for inversion is that the point (x,y) goes to \Bigl(\frac x{x^2+y^2},\frac y{x^2+y^2}\Bigr). If you substitute \frac x{x^2+y^2} for x and \frac y{x^2+y^2} for y in the equation of the ellipse, then you get \frac{\bigl(x+b(x^2+y^2)\bigr)^2}{a^2} +\frac{y^2}{b^2} = (x^2+y^2)^2. That is the equation of the limašon.
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    Your response was very helpful. I graphed your equation and it looks much more plausible.

    However, 1) (see attachments) compared to the textbook image my limacon looks disproportionately large with respect to the ellipse.

    And 2) I don't understand how you got your equation. When I substitute

    \frac{x}{x^2+y^2} for x and \frac{y}{x^2+y^2} for y in the equation \frac{(x+b)^2}{b^2}+\frac{y^2}{b^2}=1 I get:

    \frac{x^2}{x^2+y^2}+b(2x+b(x^2+y^2))+\frac{y^2}{b(  x^2+y^2)}=(x^2+y^2)<br />

    How do you arrive at \frac{(x+b(x^2+y^2))^2}{a^2}+\frac{y^2}{b^2}=(x^2+  y^2)^2 ?

    Thanks
    Attached Thumbnails Attached Thumbnails Why doesn't my ellipse inversion work gosh darn it?-ellipse-inversion-xahlee-website.bmp   Why doesn't my ellipse inversion work gosh darn it?-ellipse-inversion-attempt.jpg  
    Last edited by rainer; November 27th 2009 at 11:59 AM. Reason: mistakes in equation
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  4. #4
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    Quote Originally Posted by rainer View Post
    I graphed your equation and it looks much more plausible.

    However, 1) (see attachments) compared to the textbook image my limacon looks disproportionately large with respect to the ellipse.
    Your ellipse is a good deal smaller than the one on the text, so its inverse is larger. Your ellipse has semi-axes a=1 and b=√2/2, whereas the one in the textbook appears to have semi-axes approximately a=1.7 and b=1.3.

    Quote Originally Posted by rainer View Post
    And 2) I don't understand how you got your equation. When I substitute

    \frac{x}{x^2+y^2} for x and \frac{y}{x^2+y^2} for y in the equation \frac{x^2}{b^2}+\frac{y^2}{b^2}=1 I get something different.

    How do you arrive at \frac{(x+b(x^2+y^2))^2}{a^2}+\frac{y^2}{b^2}=(x^2+  y^2)^2 ?
    If you replace x by \frac x{x^2+y^2} then x+b becomes \frac {x + b(x^2+y^2)}{x^2+y^2}. So the equation \frac{(x+b)^2}{a^2}+\frac{y^2}{b^2} = 1 becomes \frac{\Bigl(\frac {x + b(x^2+y^2)}{x^2+y^2}\Bigr)^2}{a^2}+\frac{\Bigl(\fr  ac y{x^2+y^2}\Bigr)^2}{b^2} = 1. Multiply both sides by (x^2+y^2)^2 to get the equation of the limašon.
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  5. #5
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    Nevermind, I just realized that your equation is a more simplified form of the equation I got.

    But the limacon still looks diproportioned to me.
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    One last question-- how would I reverse the inversion (go from limacon back to ellipse)?

    Is it a simple matter of multiplying y and x by \sqrt{x^2+y^2}?
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