anonymous
  • anonymous
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.
Mathematics
chestercat
  • chestercat
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anonymous
  • 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
  • anonymous
It's also okay to express the inequalities for x in terms of y, seeing as this is a type II region.
anonymous
  • anonymous
@lgbasallote @Hero @jim_thompson5910 @dumbcow Anybody mind helping please? No one's answered for 2 hours.

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anonymous
  • anonymous
@hero is here to save the day
Hero
  • Hero
count me out
experimentX
  • experimentX
x_0 is a point, x_1 - x_2 is a distance
anonymous
  • anonymous
anonymous
  • anonymous
@experimentX x_0, x_1, x_2, y_1, y_2 are all scalars
anonymous
  • anonymous
okay, I guess I'm going to Math.SE again!
dumbcow
  • 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
  • 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
  • dumbcow
what do you mean? you have to find the area of the region
dumbcow
  • dumbcow
i thought you had to define the region with inequalities... a
anonymous
  • 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
  • experimentX
why not use piece wise functions ??
dumbcow
  • dumbcow
|dw:1346028684685:dw| the region in middle is defined as e < x < f b < y < c
anonymous
  • anonymous
That's a rectangle
dumbcow
  • dumbcow
hmm...looks like i am no help :|
anonymous
  • 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
  • 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
  • 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
  • anonymous
My suggestion would be to try polar although that has not given me any success so far.
dumbcow
  • dumbcow
we can assume the region is symmetric about x-axis correct ?
anonymous
  • anonymous
yes.
anonymous
  • anonymous
We know for a fact even
dumbcow
  • dumbcow
hmm ok im done :) last thought then is you have to break it up into different cases of possible regions
anonymous
  • anonymous
yes thats what I said you could do. But we'll see what SE says. I'll post the link soon.
anonymous
  • anonymous
good job @dumbcow
anonymous
  • anonymous
you sir, are the best
anonymous
  • anonymous
http://math.stackexchange.com/questions/187316/how-can-i-define-the-area-between-two-ellipses

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