Define the region \(\Sigma\subset\mathbb{R^2}\) as a system of inequalities for \((x, y)\). \(\Sigma\) is the intersection between the two regions defined by the ellipses (1) and (2) - see below for equations. Assume that the ellipses intersect.

- anonymous

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- anonymous

\[\frac{y^2}{y_1^2}+\frac{x^2}{x_1^2}=1\]\[\frac{y^2}{y_2^2}+\frac{(x-x_0)^2}{x_2^2}=1\]As stated in assumptions, \(x_0>x_1-x_2\), and the resulting inequalities may be expressed in terms of all subscripted constants.

- anonymous

It's also okay to express the inequalities for x in terms of y, seeing as this is a type II region.

- anonymous

@lgbasallote @Hero @jim_thompson5910 @dumbcow Anybody mind helping please? No one's answered for 2 hours.

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## More answers

- anonymous

@hero is here to save the day

- Hero

count me out

- experimentX

x_0 is a point,
x_1 - x_2 is a distance

- anonymous

@Geometry_Hater

- anonymous

@experimentX x_0, x_1, x_2, y_1, y_2 are all scalars

- anonymous

okay, I guess I'm going to Math.SE again!

- dumbcow

i would take each ellipse equation and solve for "y^2"
then set them equal in order to get it in terms of x
rearrange terms and finally solve for x_0 ...it gets messy with so many constants
then substitute that expression into inequality .... x_0 > x_1 -x_2
then rearrange terms again to solve for "x" using quadratic formula

- anonymous

Yes that would get us the two points of intersection, I agree. I can do that. But that still doesn't define the area.

- dumbcow

what do you mean? you have to find the area of the region

- dumbcow

i thought you had to define the region with inequalities... a

- anonymous

Yes. But say I have to elipses:
|dw:1346028226442:dw|
I admit that we can find a, b, and c. But how does that help us find the region in the middle?

- experimentX

why not use piece wise functions ??

- dumbcow

|dw:1346028684685:dw|
the region in middle is defined as
e < x < f
b < y < c

- anonymous

That's a rectangle

- dumbcow

hmm...looks like i am no help :|

- anonymous

It's okay. As soon as I put the question up on Math.SE I'll put up a link here if you are interested.

- dumbcow

oh i have been approaching it wrong
if you define each ellipse as function of y....f(y) and g(y)
then using your picture|dw:1346029715377:dw|
f(y) < x< g(y)
c < y < b

- anonymous

I already thought about it that way. Unfortunately, it wouldn't work:
|dw:1346029850266:dw|
Unless you want to break that up into a type II region and two type Is, that's not gonna help

- anonymous

My suggestion would be to try polar although that has not given me any success so far.

- dumbcow

we can assume the region is symmetric about x-axis correct ?

- anonymous

yes.

- anonymous

We know for a fact even

- dumbcow

hmm ok im done :)
last thought then is you have to break it up into different cases of possible regions

- anonymous

yes thats what I said you could do. But we'll see what SE says. I'll post the link soon.

- anonymous

good job @dumbcow

- anonymous

you sir, are the best

- anonymous

http://math.stackexchange.com/questions/187316/how-can-i-define-the-area-between-two-ellipses

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