6

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 :

$2ab \arctan\left(\dfrac{2ab}{(a^2 - b^2) \sin\theta}\right)$

Dheeraj Kumar
  • 2,091
Callie12
  • 581
  • 2
    Seeing as your formula was translated to LaTeX, you should look it over and confirm that the meaning wasn't changed (I was going to edit but wasn't sure what the formula meant). – theage Jan 01 '15 at 13:42
  • The area of the shape should be the sum of the areas of four elliptical sectors. An ellipse is a circle that has been stretched by a factor in one dimension, expanding its area by that factor—so if the appropriate angles can be determined, the area of such a sector can be determined from one of a circle. – G. H. Faust Jan 01 '15 at 13:49
  • I take you point but do not see a way forward yet. Thanks for your response though – Callie12 Jan 01 '15 at 14:29

1 Answers1

1

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.

Jack D'Aurizio
  • 353,855
  • Wow ! Got a couple of points which are not clear to me – Callie12 Jan 01 '15 at 17:08
  • 1
    @Callie12: which ones? – Jack D'Aurizio Jan 01 '15 at 17:16
  • Thanks Jack. If theta is angle of rotation why the theta/2 ? Also, are P1 & P2 two consecutive intersecting points of 'old' and 'rotated' ellipses ? Great stuff Jack, much appreciated – Callie12 Jan 01 '15 at 17:31
  • @Callie12: If $\theta$ is the angle between the two major axis $l,l'$, then the ellipses intersect in a point $Q_1$ such that the angle between $Q_1$ and $l$ is the same as the (opposite) angle between $Q_1$ and $l'$. For the second question, $P_1$ and $P_2$ are points on the rays emanating from the origin through the ellipses intersection points. – Jack D'Aurizio Jan 01 '15 at 17:42
  • Crystal clear Jack, I see my mistake in interpreting your solution. Apologies and thank you once again for your assistance. – Callie12 Jan 01 '15 at 17:48
  • @Callie12: you're welcome. – Jack D'Aurizio Jan 01 '15 at 17:53