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vf321
 3 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.
vf321
 3 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
 3 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
 3 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
 3 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
 3 years ago
Best ResponseYou've already chosen the best response.0@hero is here to save the day

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

vf321
 3 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
 3 years ago
Best ResponseYou've already chosen the best response.0okay, I guess I'm going to Math.SE again!

dumbcow
 3 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
 3 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
 3 years ago
Best ResponseYou've already chosen the best response.0what do you mean? you have to find the area of the region

dumbcow
 3 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
 3 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
 3 years ago
Best ResponseYou've already chosen the best response.0why not use piece wise functions ??

dumbcow
 3 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
 3 years ago
Best ResponseYou've already chosen the best response.0hmm...looks like i am no help :

vf321
 3 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
 3 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
 3 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
 3 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
 3 years ago
Best ResponseYou've already chosen the best response.0we can assume the region is symmetric about xaxis correct ?

dumbcow
 3 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
 3 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
 3 years ago
Best ResponseYou've already chosen the best response.0http://math.stackexchange.com/questions/187316/howcanidefinetheareabetweentwoellipses
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