Find the area: $(\frac xa+\frac yb)^2 = \frac xa-\frac yb , y=0 , b>a$ Find the area: $$\left(\frac xa+\frac yb\right)^2 = \frac xa-\frac yb,$$ $ y=0 , b>a$
I work in spherical coordinates. 
$x=a\cdot r\cdot \cos(\phi)\;\;,$
$y=b\cdot r\cdot \cos(\phi)$
Then I get the equation and don't know to do with, cause "a" and "b" are dissapearing.
For what are the conditions: $y=0, b>a?$..How to define the limits of integration?

$X=\frac{x}{a}=r\cos \theta\\ Y=\frac{y}{b}=r\sin \theta$
The curve becomes $(X+Y)^2=X-Y$ or $R^2(cos^2\theta+sin^2\theta+2\cos \theta \sin \theta)=R(\cos \theta -\sin \theta)$ which becomes $R=\displaystyle{\frac{\cos \theta -\sin \theta}{1+2\cos \theta \sin \theta}}$.
area$\displaystyle{=\int_0^{\frac{\tau}{2}=\pi}\int_0^R}ab \ \mathrm{d}r\mathrm{d}\theta= ab \int_0^{\frac{\tau}{2}}R \ \mathrm{d}\theta= ab \left[ \frac{-1}{\sin \theta + \cos \theta }\right]_0^{\frac{\tau}{2}}=2ab$
