# Angle of an ellipse point shifted up in respect to a similar at origin

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.

• 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. Jan 14 '21 at 21:53
• 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. Jan 14 '21 at 21:56
• @DavidK Right, I edited it, so with a condition with $h < b$. Jan 14 '21 at 22:06
• @Intelligentipauca Yeah, I forgot to edit it, thanks. I fixed it. Jan 14 '21 at 22:06

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 $$(2)$$.