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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- Sep 13th 2012, 05:08 PMZeroDivisionErrorDiagonal 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.

- Sep 13th 2012, 05:58 PMjohnsomeoneRe: Diagonal ellipse
$\displaystyle ax^2 + bxy + cy^2 + dx + ey + f = 0$ is the equation for the general "rotated & shifted" conic section in the plane.

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

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. - Sep 18th 2012, 08:10 AMZeroDivisionErrorRe: Diagonal ellipse
Thank you so much! I will definitely print this out. Love you, stranger! (In a platonic way.)

- Sep 18th 2012, 08:37 AMjohnsomeoneRe: 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. - Sep 18th 2012, 12:05 PMHallsofIvyRe: Diagonal ellipse
The general formula for an ellipse with center at (a, b) and axes parallel to the x and y axes is $\displaystyle \frac{x^2}{a^2}+\frac{y^2}{b^2}= 1$ or $\displaystyle b^2x^2+ a^2y^2= a^2b^2$. To rotate so that the axis parallel to the x-axis moves to angle $\displaystyle \theta$ with the x-axis, substitute $\displaystyle cos(\theta)+ ysin(\theta)$ for x and $\displaystyle -y'sin(\theta)+ x'cos(\theta)$ for y.