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Any arbitrary ellipse in the x-y plane can be described with five parameters -- usually the center’s x and y coordinate positions, x0 and y0; the distance between focal points, d; the eccentricity, $\epsilon$; and the counterclockwise tilt angle, $\psi$, of the line through the center and foci with respect to the horizontal x axis. There are some trivially derived variants of this set of five, such as using the major and minor semi-axes or doing a simple transformation to polar coordinates, but I am trying to find analytic expressions for x0 and y0 if they are not given but rather the coordinates (x1, y1) of some particular point on the perimeter are known instead. In other words I want to know if there is an analytic closed form expression for F and G where

 x0 = F(d,$\epsilon$,$\psi$,x1,y1)
 y0 = G(d,$\epsilon$,$\psi$,x1,y1)

If the counterclockwise angle $\theta$ of the tangent line through the perimeter point (x1, y1) with respect to the horizontal x axis is also given or known could that be sufficient to determine the tilt angle $\psi$ if it is not known explicitly?

Follow up

Thank you J. M. and Barry Cipra. I see the point you are making, and think that I need to explain the context. I have a software program that, when given

(1) the coordinates of a starting perimeter point, (x1, y1);
(2) a tilt angle, $\psi$ (3) a distance between foci, d (4) an eccentricity, $\epsilon$ (5) a starting perimeter point tangential angle, $\theta$ (6) a travel angle, $\phi$, with respect to a line from a center through the starting point

it will generate an elliptical arc segment whose end point is determined by these inputs. However there are times when I want to be able to specify the coordinates of the end point, (x2, y2), explicitly as input and manipulate the shape of the arc between the two fixed points with the other input parameters. I think I could modify the program for this if there were some way to express the coordinate positions of the center of the ellipse in terms of these input parameters.

Because an ellipse has five degrees of freedom for parameters, if I specify the four position coordinates of the start and end points of an arc segment (x1, y1, x2, y2), I think that specifying one further parameter from ($\psi$, d, $\epsilon$, $\theta$, or $\phi$) ought to determine the ellipse and consequently determine the other four unspecified ones among these, as well as determining expressions for other commonly used ellipse parameters (center coordinates x0 and y0, semi-major and semi-minor axes, coordinates of foci, etc.) as functions of the five specified ones.

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Only one perimeter point? Your problem's ill-posed. I can draw any number of congruent ellipses with axes perpendicular to the coordinate axes, and passing through the origin, to use an easy example. – J. M. Nov 22 '11 at 7:43
up vote 3 down vote accepted

To elaborate on J.M.'s comment, your parameters $d$ and $\epsilon$ determine the size and shape of an ellipse, and $\psi$ determines its orientation in the plane, but once those are set you're still free to rigidly translate the ellipse (without rotating it), which means the point $(x_1,y_1)$ can be anywhere along its perimeter. If you replace $\psi$ with $\theta$, you're now allowing yourself to rotate the ellipse and then place it up against a specified tangent line. (If you keep $\psi$ and add $\theta$, you've got it down to two possibilities, one on either side of the tangent line, but at this point you've got six parameters instead of five.) So all in all, the answer to your question is no, there's no way to express $(x_0,y_0)$ uniquely in terms of $d$, $\epsilon$, $\psi$ (or $\theta$), $x_1$ and $y_1$.

You might, however, be able to solve for the ellipse whose center is closest to the origin. But that's a different question.

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Oh damn! The reputation points I've gotten for this answer have shot me past my goal. I was hoping to hit and then linger for a while at 666.... – Barry Cipra Nov 22 '11 at 21:42
Ask a question, assign a carefully computed bounty, accept the answer and you'll have a second chance :) – fedja Nov 22 '11 at 22:17
@fedja, thanks, that's a hell of a good idea. Maybe I can ask something about the Devil's Staircase. – Barry Cipra Nov 22 '11 at 22:25

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