## anonymous 4 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.

1. 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.

2. anonymous

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

3. anonymous

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

4. anonymous

@hero is here to save the day

5. Hero

count me out

6. experimentX

x_0 is a point, x_1 - x_2 is a distance

7. anonymous

@Geometry_Hater

8. anonymous

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

9. anonymous

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

10. anonymous

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

11. 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.

12. anonymous

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

13. anonymous

i thought you had to define the region with inequalities... a<x<b and c<y<d is that right

14. 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?

15. experimentX

why not use piece wise functions ??

16. anonymous

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

17. anonymous

That's a rectangle

18. anonymous

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

19. 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.

20. anonymous

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

21. 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

22. anonymous

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

23. anonymous

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

24. anonymous

yes.

25. anonymous

We know for a fact even

26. anonymous

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

27. anonymous

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

28. anonymous

good job @dumbcow

29. anonymous

you sir, are the best

30. anonymous