Given an ellipse with dimensions 165 x 135, how can I find the angle α as shown in the image? All distances from the blue circular arc to point M are 50 units.
Using the equation $F=\sqrt{(y/2)^2-(x/2)^2}$ I found the distance between foci points and the centre of the ellipse to be 47.43, however I am not sure how to appropiately use this value.
Let $H$ be the projection of $P$ on the major axis and $O$ the ellipse center. Then: $$ PH=D\sin\alpha,\quad OH=a-D\cos\alpha, $$ where $a=165/2$ m is the semi major axis. Plug these into the ellipse equation: $OH^2/a^2+PH^2/b^2=1$ (where $b=135/2$ m is the semi minor axis) to get an equation for $\cos\alpha$: $$ {D^2(a^2-b^2)\over a^2b^2}\cos^2\alpha+2{D\over a}\cos\alpha-{D^2\over b^2}=0. $$