Given two ellipses that take up regions $E_1$ and $E_2$ in $\mathbb{R^2}$, with the following properties:
- Centers defined in the Cartesian coordinate system $(c_1, 0)$ for $E_1$ and $(c_2, 0)$ for $E_2$ such that $c_2>c_1$
- Semi-diameters $x_1$ & $y_1$ for $E_1$ and $x_2$ & $y_2$ for $E_2$ such that the two ellipses intersect at exactly two points.
Let the surface $\Sigma=E_1\cap E_2$
Describe $\Sigma$ with two parameters, $u$ and $v$, in the form $a\le u\le b$ and $f(u)\le v \le g(u)$ for $\left\{a,b:a,b\in\mathbb{R}\land a, $f:\mathbb{R}\rightarrow\mathbb{R}$, $g:\mathbb{R}\rightarrow\mathbb{R}$, and $g(u)>f(u)$ over $[a,b]$. In other words, please find the Lebesgue measure of the set of points that satisfy both of the aforementioned inequalities.
-My work so far-
The two equations for the ellipses are easy to find given the conditions. $ \frac{y^2}{y_1^2}+\frac{(x-c_1)^2}{x_1^2}=1 $ $ \frac{y^2}{y_2^2}+\frac{(x-c_2)^2}{x_2^2}=1 $ After multiplying by a constant and subtracting one from another, I arrive at two solutions for $x$: $ x=\frac{b\pm \sqrt{b^2-4ac}}{2a} $ Where $a=\frac{y_1^2}{x_1^2}-\frac{y_2^2}{x_2^2}$, $b=\frac{y_2^2c_2}{x_2^2}-\frac{y_1^2c_1}{x_1^2}$, and $c=\frac{y_1^2c_1^2}{x_1^2}+\frac{y_2^2c_2^2}{x_2^2}$. However, from the fact that $c_2>c_1$ and that the ellipses only intersect at two points, I suppose that the only value of $x$ must be the larger one. I evaluated and checked the determinant with Mathematica, and it is not equal to 0. How can I be sure to pick the right value of $x$?
Even assuming I found the right value of $x$ and therefore have the intersection points $(x_0, y_0)$ and $(x_0, -y_0)$, with $y_0=\sqrt{y_2^2(1-\frac{(x_0-c_2)^2}{x_1^2})}$, I still have the problem of defining the intersecting area.
If I have $u=y$ and $v=x(y)$, then I'm assuming I have a type II region, whereas I really have a type II region combined with two type I regions. Pursuing that piecewise definition of $\Sigma$ is not optimal, since I would have to find what parts of the two curves break the horizontal line test, set up three different integrals, etc. A potential workaround I see is converting to polar, but I don't know how to approach that.
So, to summarize:
- If you could find a way to solve the original, that would be much appreciated.
- If you could tell me which $x$ value in the quadratic equation is $x_0$, please let me know.
- If you could elaborate a polar approach, I'd love to see it.
EDIT
As pointed out in the comments section below, the value $x_0$ must be only one of the following solutions $x=\frac{b\pm \sqrt{b^2-4ac}}{2a}$, since the $x$ value farther from $c_2$ will yield an imaginary $y_0$ value. However, this means I cannot find $x$ symbolically. If someone has a different approach to finding the intersection, please let me know.