I would like to get the topmost (bottommost depending on the curve) point of an arc which means the point where the tangent is 0.
so we know that $y = c_y + r_y\sin(\alpha)\cos(t) + r_x\cos(\alpha)\sin(t)$ where $\alpha$ is the x-axis rotation value of the ellipse
we know that the tangent gives $dy/dt = 0$ (if I am not mistaken) which mean $-r_y\sin(\alpha)\sin(t) + r_x\cos(\alpha)\cos(t) = 0$
which gives $$t = \arctan{r_x\over r_y\tan(\alpha)}$$ of course $\cos(t)$ should be different from 0
but this doesn't seem to be the right result
here is desmos the A value is alpha
so first we have $y = c_y + r_x\sin(\alpha)\cos(t) + r_y\cos(\alpha)\sin(t)$
which gives $$t = \arctan{r_y\over r_x\tan(\alpha)}$$
after that, we should calculate the following (unrotated ellipse equation)
$u=a\cdot\cos\left(\arctan\left(\frac{b}{a\cdot\tan\left(\alpha\right)}\right)\right)$
$v=b\cdot\sin\left(\arctan\left(\frac{b}{a\cdot\tan\left(\alpha\right)}\right)\right)$
and lastly, we rotate the point (u, v) by $\alpha$
$x=u\cdot\cos\left(\alpha\right)-v\cdot\sin\left(\alpha\right)$
$y=u\cdot\sin\left(\alpha\right)+v\cdot\cos\left(\alpha\right)$
the wanted point is (x, y)
desmos