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vf321
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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.
 one year ago
 one year ago
vf321 Group Title
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.
 one year ago
 one year ago

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vf321 Group TitleBest ResponseYou've already chosen the best response.0
\[\frac{y^2}{y_1^2}+\frac{x^2}{x_1^2}=1\]\[\frac{y^2}{y_2^2}+\frac{(xx_0)^2}{x_2^2}=1\]As stated in assumptions, \(x_0>x_1x_2\), and the resulting inequalities may be expressed in terms of all subscripted constants.
 one year ago

vf321 Group TitleBest ResponseYou've already chosen the best response.0
It's also okay to express the inequalities for x in terms of y, seeing as this is a type II region.
 one year ago

vf321 Group TitleBest ResponseYou've already chosen the best response.0
@lgbasallote @Hero @jim_thompson5910 @dumbcow Anybody mind helping please? No one's answered for 2 hours.
 one year ago

panlac01 Group TitleBest ResponseYou've already chosen the best response.0
@hero is here to save the day
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.0
x_0 is a point, x_1  x_2 is a distance
 one year ago

panlac01 Group TitleBest ResponseYou've already chosen the best response.0
@Geometry_Hater
 one year ago

vf321 Group TitleBest ResponseYou've already chosen the best response.0
@experimentX x_0, x_1, x_2, y_1, y_2 are all scalars
 one year ago

vf321 Group TitleBest ResponseYou've already chosen the best response.0
okay, I guess I'm going to Math.SE again!
 one year ago

dumbcow Group TitleBest ResponseYou've already chosen the best response.0
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
 one year ago

vf321 Group TitleBest ResponseYou've already chosen the best response.0
Yes that would get us the two points of intersection, I agree. I can do that. But that still doesn't define the area.
 one year ago

dumbcow Group TitleBest ResponseYou've already chosen the best response.0
what do you mean? you have to find the area of the region
 one year ago

dumbcow Group TitleBest ResponseYou've already chosen the best response.0
i thought you had to define the region with inequalities... a<x<b and c<y<d is that right
 one year ago

vf321 Group TitleBest ResponseYou've already chosen the best response.0
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?
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.0
why not use piece wise functions ??
 one year ago

dumbcow Group TitleBest ResponseYou've already chosen the best response.0
dw:1346028684685:dw the region in middle is defined as e < x < f b < y < c
 one year ago

vf321 Group TitleBest ResponseYou've already chosen the best response.0
That's a rectangle
 one year ago

dumbcow Group TitleBest ResponseYou've already chosen the best response.0
hmm...looks like i am no help :
 one year ago

vf321 Group TitleBest ResponseYou've already chosen the best response.0
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.
 one year ago

dumbcow Group TitleBest ResponseYou've already chosen the best response.0
oh i have been approaching it wrong if you define each ellipse as function of y....f(y) and g(y) then using your picturedw:1346029715377:dw f(y) < x< g(y) c < y < b
 one year ago

vf321 Group TitleBest ResponseYou've already chosen the best response.0
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
 one year ago

vf321 Group TitleBest ResponseYou've already chosen the best response.0
My suggestion would be to try polar although that has not given me any success so far.
 one year ago

dumbcow Group TitleBest ResponseYou've already chosen the best response.0
we can assume the region is symmetric about xaxis correct ?
 one year ago

vf321 Group TitleBest ResponseYou've already chosen the best response.0
We know for a fact even
 one year ago

dumbcow Group TitleBest ResponseYou've already chosen the best response.0
hmm ok im done :) last thought then is you have to break it up into different cases of possible regions
 one year ago

vf321 Group TitleBest ResponseYou've already chosen the best response.0
yes thats what I said you could do. But we'll see what SE says. I'll post the link soon.
 one year ago

panlac01 Group TitleBest ResponseYou've already chosen the best response.0
good job @dumbcow
 one year ago

panlac01 Group TitleBest ResponseYou've already chosen the best response.0
you sir, are the best
 one year ago

vf321 Group TitleBest ResponseYou've already chosen the best response.0
http://math.stackexchange.com/questions/187316/howcanidefinetheareabetweentwoellipses
 one year ago
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