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I have an ellipse $(E): x = a\cdot \cos(\theta), y = b\cdot \sin(\theta)$ then I move this ellipse upwards along the y-axis (semi-minor axis) so the ellipse $E$ become an ellipse $E'$ (in green) not centered at origin $O(0,0)$ but at $O'(0,h)$ with $h < b $.

How to calculate the angle $\beta$, the new angle between the semi-axis major and point $A$ on $E'$ in terms of $\alpha$, the old angle between the semi-axis major and $A$ when $E'$ was centered at $O$.?

I tried something with eccentric angle but the shifting here is not along the foci axis.

I tried and found a relation of $\alpha$ in terms of $\beta$ but not my case thus the formula was irreversible. Any tips? thanks in advanve.

Please see the illustration illustration.

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    $\begingroup$ If you move the ellipse farther than the semi-minor axis, then a ray in any given direction $\alpha$ from the origin will almost always intersect the ellipse twice (two values of $\beta$) or not at all (no values of $\beta$). The only place where you could deduce $\beta$ uniquely from $\alpha$ would be on two lines through the origin tangent to the ellipse. $\endgroup$
    – David K
    Jan 14 '21 at 21:53
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    $\begingroup$ Note that angle $\alpha$ in the figure is not the same as the parameter in the equations of the ellipse. You'd better choose another name for the parameter. $\endgroup$ Jan 14 '21 at 21:56
  • $\begingroup$ @DavidK Right, I edited it, so with a condition with $h < b$. $\endgroup$
    – Khaled
    Jan 14 '21 at 22:06
  • $\begingroup$ @Intelligentipauca Yeah, I forgot to edit it, thanks. I fixed it. $\endgroup$
    – Khaled
    Jan 14 '21 at 22:06
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If you just need to compute the answer for any given input, consider the equation of the line $OA,$ which is $$y = x\tan\alpha.\tag1$$ You also have an equation of the ellipse in the form $$ \frac{x^2}{a^2} + \frac{(y - h)^2}{b^2} = 1. \tag2 $$ Taking these as simultaneous equations, solve for $x$ and $y$. There are two solutions, one of which is the coordinates of $A$. The coordinates of $A$ will be the ones where the $x$-coordinate has the same sign as $\cos\alpha.$

Another approach is to transform the ellipse to a circle. Multiply the $y$ coordinate of every point by $\frac ba.$ This maps the ellipse to a circle of radius $a$ with its center at $O'' = (0,h')$ where $h' = \frac ab h$ and maps the point $A = (x_A,y_A)$ to $A' = (x_A,y'_A)$ where $y'_A = \frac ab y_A.$ Instead of angles $\alpha$ and $\beta$ you have angles $\alpha'$ and $\beta'$ where $\tan\alpha' = \frac ab \tan\alpha$ and $\tan\beta' = \frac ab \tan\beta.$ The problem is simpler for the circle than the ellipse, and you can work out $\beta'$ in terms of $\alpha'$ with some trigonometry, though if you then express $\alpha'$ in terms of $\alpha$ and $\beta$ in terms of $\beta'$ in order to have a formula for $\beta$ in terms of $\alpha,$ the formula gets quite long and complicated.

If I were programming this I think I would prefer solving Equations $(1)$ and $(2)$.

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  • $\begingroup$ Yes, that is! thank you very much. I toke your first approach then I take in account the real quadrant of angles. $\endgroup$
    – Khaled
    Jan 15 '21 at 8:51

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