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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.

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\[\frac{y^2}{y_1^2}+\frac{x^2}{x_1^2}=1\]\[\frac{y^2}{y_2^2}+\frac{(x-x_0)^2}{x_2^2}=1\]As stated in assumptions, \(x_0>x_1-x_2\), and the resulting inequalities may be expressed in terms of all subscripted constants.
It's also okay to express the inequalities for x in terms of y, seeing as this is a type II region.
@lgbasallote @Hero @jim_thompson5910 @dumbcow Anybody mind helping please? No one's answered for 2 hours.

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@hero is here to save the day
count me out
x_0 is a point, x_1 - x_2 is a distance
@experimentX x_0, x_1, x_2, y_1, y_2 are all scalars
okay, I guess I'm going to Math.SE again!
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 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
Yes that would get us the two points of intersection, I agree. I can do that. But that still doesn't define the area.
what do you mean? you have to find the area of the region
i thought you had to define the region with inequalities... a
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?
why not use piece wise functions ??
|dw:1346028684685:dw| the region in middle is defined as e < x < f b < y < c
That's a rectangle
hmm...looks like i am no help :|
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.
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
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
My suggestion would be to try polar although that has not given me any success so far.
we can assume the region is symmetric about x-axis correct ?
We know for a fact even
hmm ok im done :) last thought then is you have to break it up into different cases of possible regions
yes thats what I said you could do. But we'll see what SE says. I'll post the link soon.
good job @dumbcow
you sir, are the best

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