area of ellipse in different quadrants

Question

Let the equation of the ellipse $\displaystyle \frac{(x-20)^2}{20}+\frac{(y-16)^2}{16}=2016$. Let $R_1$, $R_2$, $R_3$, $R_4$ denote the area of the ellipse in first, second, third and fourth quadrants respectively, then $R_1-R_2+R_3-R_4$ equals—

From the problem it is clear that the major axis of the ellipse is $\sqrt{20\times2016}$ units and the minor axis is $4\sqrt{2016}$ units. The centre of the ellipse is $(20,16)$ and too bad, the ellipse is not symmetrical about any axis. I tried solving this question using integration but the integration too seemed too cumbersome and lengthy. This is a question asked in some entrance test hence it should have a fast solution indeed. Please help me out. Many thanks!

Answer 1

Considering $$\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1$$

where $h,k>0$.

Let $x_{+}$, $x_{-}$ be the $x$-intercepts and $y_{+}$, $y_{-}$ be the $y$-intercepts

where $x_{-}<0<x_{+}$, $y_{-}<0<y_{+}$

By doing simple dissection of the ellipse below:

The red and blue regions below are confined in the region of $0<x<x_{+}+x_{-}$

The purple region below is confined in the region of $0<y<y_{+}+y_{-}$

Hence the required area is \begin{align*} R_{1}-R_{2}+R_{3}-R_{4} &= (x_{+}+x_{-})(y_{+}+y_{-}) \\ &=(2h)(2k) \\ &= 4hk \end{align*}

In general, it holds for any $h$ and $k$ such that $\displaystyle \frac{h^2}{a^2}+\frac{k^2}{b^2}<1$.