Given the following image (not quite to scale, the red lines for h should form a square):
I need to find the length of the semi-major axis of this ellipse, notated on the diagram as w.
The semi-minor axis h and angle θ are known values. The side of the enclosing rectangle is also distance h from the center of the ellipse. The angle θ measures the angle to the ellipse tangent line at the point of intersection between the ellipse and enclosing rectangle.
I've tried various methods of getting this length w using a combination of trig and calculus, but I always end up with rational equations that I am not sure how to solve. Moreover, I'm not certain that my approach is even close to correct. So all that said:
How can I create a function f(h, θ) which produces the value w?
For those wondering about the motivation here: The enclosing rectangle is a phone screen. The top arc of the ellipse forms a curved border between page elements, and we want the apparent angle at which that border meets the screen edge to remain constant for all device widths.
Parametrize the ellipse as $$(x(t), y(t)) = (w \cos t, h \sin t), \quad 0 \le t < 2\pi, \quad w > h > 0. \tag{1}$$ Then the point of tangency corresponds to a value of $t$, say $t = t_0$, such that $x(t_0) = h$, or $$t_0 = \arccos \frac{h}{w}. \tag{2}$$ The slope $m(t)$ of the tangent line at $t = t_0$ is given by $$m(t_0) = \left[\frac{dy/dt}{dx/dt}\right]_{t=t_0} = \left[-\frac{h}{w} \cot t \right]_{t=t_0} = -\frac{h^2}{w \sqrt{w^2 - h^2}}. \tag{3}$$ The angle of the tangent line is the arctangent of the slope; i.e. $$\tan \theta = m(t_0). \tag{4}$$ So to obtain the semimajor axis $w$ in terms of a given fixed semiminor axis $h$ and desired tangent angle $\theta$, we solve the equation $(4)$ after substituting in $(3)$, giving $$w = \frac{h}{\sqrt{2}} \sqrt{1 + \sqrt{1 + 4 \cot^2 \theta}}.$$