- 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.

- chestercat

See more answers at brainly.com

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions

- 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

It's also okay to express the inequalities for x in terms of y, seeing as this is a type II region.

- anonymous

@lgbasallote @Hero @jim_thompson5910 @dumbcow Anybody mind helping please? No one's answered for 2 hours.

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- anonymous

@hero is here to save the day

- Hero

count me out

- experimentX

x_0 is a point,
x_1 - x_2 is a distance

- anonymous

- anonymous

@experimentX x_0, x_1, x_2, y_1, y_2 are all scalars

- anonymous

okay, I guess I'm going to Math.SE again!

- 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

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

what do you mean? you have to find the area of the region

- dumbcow

i thought you had to define the region with inequalities... a

- 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

why not use piece wise functions ??

- dumbcow

|dw:1346028684685:dw|
the region in middle is defined as
e < x < f
b < y < c

- anonymous

That's a rectangle

- dumbcow

hmm...looks like i am no help :|

- 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

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

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

My suggestion would be to try polar although that has not given me any success so far.

- dumbcow

we can assume the region is symmetric about x-axis correct ?

- anonymous

yes.

- anonymous

We know for a fact even

- dumbcow

hmm ok im done :)
last thought then is you have to break it up into different cases of possible regions

- anonymous

yes thats what I said you could do. But we'll see what SE says. I'll post the link soon.

- anonymous

good job @dumbcow

- anonymous

you sir, are the best

- anonymous

http://math.stackexchange.com/questions/187316/how-can-i-define-the-area-between-two-ellipses

Looking for something else?

Not the answer you are looking for? Search for more explanations.