I am given an ellipse with known semi-major axis $a$ and semi-minor axis $b$. The ellipse is placed in a rectangle of horizontal width $w$ such that $ 2 b \le w \le 2 a $, as shown in the attached figure. Find the angle $\alpha$ that the major axis makes with the positive $x$ axis.
From symmetry, the center of the ellipse is at
$ C = (C_x, C_y) $
with $C_x = \dfrac{w}{2} $ and $C_y$ is unknown.
Let the vectors along the semi-major axis and semi-minor axes be $v_1$ and $v_2$ respectively, then the vector equation of the ellipse is
$ r(t) = C + v_1 \cos(t) + v_2 \sin(t) $
where
$ v_1 = (a \cos(\alpha), a \sin(\alpha) ) , v_2 = (- b \sin(\alpha), b \cos(\alpha) ) $
It follows that the $x$-coordinate of $r(t)$ is given by
$ x(t) = \dfrac{w}{2} + a \cos(\alpha) \cos(t) - b \sin(\alpha) \sin(t) $
From the diagram, the maximum $x$ is $w$. Hence
$ w = \dfrac{w}{2} + \sqrt{ a^2 \cos^2(\alpha) + b^2 \sin^2(\alpha) } $
So that
$ \dfrac{w^2}{4} = \big(\dfrac{ a^2 + b^2 }{2}\big) + \big(\dfrac{a^2 - b^2}{2}\big) \cos(2 \alpha) $
Now we can easily solve for $\alpha$.