Area common to two equal ellipses with same centre but axes rotated by an acute angle Been some time since I did this kind of geometry and it seems to have me stumped already but looks so innocent. Any help would be most welcomed.
The answer is quoted to be :
$2ab \arctan\left(\dfrac{2ab}{(a^2 - b^2) \sin\theta}\right)$
 A: As pointed out by G.H. Faust, we just need to compute four times the area of the elliptic sector between the angles $\frac{\theta}{2}$ and $\frac{\pi+\theta}{2}$. By multiplying the $x$-coordinate by $b$ and the $y$-coordinate by $a$, the ellipse is mapped into a circle having radius $ab$, and the vertices of the previous elliptic sector fall in
$$P_1=\nu\cdot\left(b\cos\frac{\theta}{2},a\sin\frac{\theta}{2}\right),\quad P_2=\eta\cdot\left(-b\sin\frac{\theta}{2},a\cos\frac{\theta}{2}\right).$$
We have:
$$\arg P_1 = \arctan\left(\frac{a}{b}\tan\frac{\theta}{2}\right),\qquad \arg P_2=\frac{\pi}{2}+\arctan\left(\frac{b}{a}\tan\frac{\theta}{2}\right)$$
hence:
$$\widehat{P_1 O P_2}=\frac{\pi}{2}+\arctan\left(\frac{\frac{b}{a}\tan\frac{\theta}{2}-\frac{a}{b}\tan\frac{\theta}{2}}{1+\tan^2\frac{\theta}{2}}\right)=\frac{\pi}{2}+\arctan\left(\frac{\sin\theta}{2}\left(\frac{b}{a}-\frac{a}{b}\right)\right)$$
or:
$$\widehat{P_1 O P_2}=\arctan\left(\frac{2ab}{(a^2-b^2)\sin\theta}\right)$$
and by mapping back the circular sector given by $P_1,O,P_2$ into the elliptic sector, we get that the area of the latter is
$$\frac{ab}{2}\arctan\left(\frac{2ab}{(a^2-b^2)\sin\theta}\right)$$
as stated.
