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

geometrypolar coordinates

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.

Best Answer

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 $(2)$.