A book that I've read shows how to find the area of the shadow cast by a sphere and ellipsoid. The spherical shadow makes sense; its simply the area of a circle (which would be the sphere's shadow) divided by $\sin\beta$ (where $\beta$ is the sun's elevation angle, $90^\circ$ at noon). This makes sense to me since when the sun is directly over the sphere, it's like looking at it as if it was a circle instead, so the shadow cast would just be the simple circle.
I thought I could employ the same logic for a rectangular plane; if we have a floating rectangle of length $l$ and width $w$, it would have an area of $A_r=lw$, and its shadow would have the area $$A_s=\frac{lw}{\sin\beta}\ .$$ However, this apparently isn't the case! Take for example the shadow cast by an ellipsoid; according to the book, even though the area of an ellipse is $A_e=\pi ab$, the shadow cast by the ellipsoid is given by some convoluted formula $$A_h=\pi b^2\left(1+\frac{a^2}{b^2\tan^2\beta}\right)\ ,$$ whereas I thought it would've been similar to the area of the sphere's shadow, $$A_h=\frac{\pi ab}{\sin\beta}\ .$$ Now, I know that shadows get distorted as the sun rises and sets, and depending on your location in the world (north pole vs equator, and also day of year). Ignoring that, however, and simply focusing on the time of day (i.e. sun at angle $\beta$), how can I find the floating (parallel to the ground) rectangular plane's shadow length and width? Note that the sun can be at any point $(x,y)$ -- it doesn't have to simply move along the $x$ or $y$ axis, meaning both the shadow's length and width can change! I've been trying to tackle this for days with no luck :(
"Bonus" (less important than original question, but if someone could help me on this also I'd appreciate it), assume the floating rectangular plane is always perpendicular to the sun (and moves itself to remain that way throughout the day), and is thus not always parallel to the ground (as before); how would this change the shadow length and width?