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 2 years ago
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.
 2 years ago
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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vf321
 2 years ago
Best 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.

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

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

panlac01
 2 years ago
Best ResponseYou've already chosen the best response.0@hero is here to save the day

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.0x_0 is a point, x_1  x_2 is a distance

vf321
 2 years ago
Best ResponseYou've already chosen the best response.0@experimentX x_0, x_1, x_2, y_1, y_2 are all scalars

vf321
 2 years ago
Best ResponseYou've already chosen the best response.0okay, I guess I'm going to Math.SE again!

dumbcow
 2 years ago
Best ResponseYou've already chosen the best response.0i 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

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

dumbcow
 2 years ago
Best ResponseYou've already chosen the best response.0what do you mean? you have to find the area of the region

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

vf321
 2 years ago
Best ResponseYou've already chosen the best response.0Yes. 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
 2 years ago
Best ResponseYou've already chosen the best response.0why not use piece wise functions ??

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

dumbcow
 2 years ago
Best ResponseYou've already chosen the best response.0hmm...looks like i am no help :

vf321
 2 years ago
Best ResponseYou've already chosen the best response.0It'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
 2 years ago
Best ResponseYou've already chosen the best response.0oh 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

vf321
 2 years ago
Best ResponseYou've already chosen the best response.0I 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

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

dumbcow
 2 years ago
Best ResponseYou've already chosen the best response.0we can assume the region is symmetric about xaxis correct ?

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

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

vf321
 2 years ago
Best ResponseYou've already chosen the best response.0http://math.stackexchange.com/questions/187316/howcanidefinetheareabetweentwoellipses
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