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Math Help - Diagonal ellipse

  1. #1
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    Diagonal ellipse

    Does anyone know a way to graph an ellipse such that the slope of the line segment between the foci is nonzero and also non-infinite? (Basically, does anyone know a Cartesian equation for a diagonal ellipse?) Parametric works.
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  2. #2
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    Re: Diagonal ellipse

    ax^2 + bxy + cy^2 + dx + ey + f = 0 is the equation for the general "rotated & shifted" conic section in the plane.

    If D = b^2- 4ac, then it's an ellipse for D<0, a parabola for D = 0, and a hyperbola for D>0.

    The a, b, c give the general shape & type (via D), and then the e and f give the translation. The bxy term is what appears when the conics are rotated from their usual positions nicely aligned to the x & y axes.
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  3. #3
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    Re: Diagonal ellipse

    Thank you so much! I will definitely print this out. Love you, stranger! (In a platonic way.)
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  4. #4
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    Re: Diagonal ellipse

    Glad to help.

    If you're going to print it out, I should point out a tiny error I now see in what I wrote. My original "e and f give the translation" was incorrect. It shoud read "e and d give the translation."

    Also, that doesn't mean that you can read off the x and y translations from e and d - the relationship is more complicated. It only means that when those terms show up, it indicates that the conic section is no longer centered at the origin.

    Good luck with your studies.
    Last edited by johnsomeone; September 18th 2012 at 08:42 AM.
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  5. #5
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    Re: Diagonal ellipse

    The general formula for an ellipse with center at (a, b) and axes parallel to the x and y axes is \frac{x^2}{a^2}+\frac{y^2}{b^2}= 1 or b^2x^2+ a^2y^2= a^2b^2. To rotate so that the axis parallel to the x-axis moves to angle \theta with the x-axis, substitute cos(\theta)+ ysin(\theta) for x and -y'sin(\theta)+ x'cos(\theta) for y.
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