## 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

1. vf321 Group Title

$\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. vf321 Group Title

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

3. vf321 Group Title

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

4. panlac01 Group Title

@hero is here to save the day

5. Hero Group Title

count me out

6. experimentX Group Title

x_0 is a point, x_1 - x_2 is a distance

7. panlac01 Group Title

@Geometry_Hater

8. vf321 Group Title

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

9. vf321 Group Title

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

10. dumbcow Group Title

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. vf321 Group Title

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. dumbcow Group Title

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

13. dumbcow Group Title

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

14. vf321 Group Title

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 Group Title

why not use piece wise functions ??

16. dumbcow Group Title

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

17. vf321 Group Title

That's a rectangle

18. dumbcow Group Title

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

19. vf321 Group Title

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. dumbcow Group Title

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. vf321 Group Title

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. vf321 Group Title

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

23. dumbcow Group Title

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

24. vf321 Group Title

yes.

25. vf321 Group Title

We know for a fact even

26. dumbcow Group Title

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

27. vf321 Group Title

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

28. panlac01 Group Title

good job @dumbcow

29. panlac01 Group Title

you sir, are the best

30. vf321 Group Title