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vf321

  • 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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  1. vf321
    • 2 years ago
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    \[\frac{y^2}{y_1^2}+\frac{x^2}{x_1^2}=1\]\[\frac{y^2}{y_2^2}+\frac{(x-x_0)^2}{x_2^2}=1\]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
    • 2 years ago
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    It's also okay to express the inequalities for x in terms of y, seeing as this is a type II region.

  3. vf321
    • 2 years ago
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    @lgbasallote @Hero @jim_thompson5910 @dumbcow Anybody mind helping please? No one's answered for 2 hours.

  4. panlac01
    • 2 years ago
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    @hero is here to save the day

  5. Hero
    • 2 years ago
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    count me out

  6. experimentX
    • 2 years ago
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    x_0 is a point, x_1 - x_2 is a distance

  7. panlac01
    • 2 years ago
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    @Geometry_Hater

  8. vf321
    • 2 years ago
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    @experimentX x_0, x_1, x_2, y_1, y_2 are all scalars

  9. vf321
    • 2 years ago
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    okay, I guess I'm going to Math.SE again!

  10. dumbcow
    • 2 years ago
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    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
    • 2 years ago
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    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
    • 2 years ago
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    what do you mean? you have to find the area of the region

  13. dumbcow
    • 2 years ago
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    i thought you had to define the region with inequalities... a<x<b and c<y<d is that right

  14. vf321
    • 2 years ago
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    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
    • 2 years ago
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    why not use piece wise functions ??

  16. dumbcow
    • 2 years ago
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    |dw:1346028684685:dw| the region in middle is defined as e < x < f b < y < c

  17. vf321
    • 2 years ago
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    That's a rectangle

  18. dumbcow
    • 2 years ago
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    hmm...looks like i am no help :|

  19. vf321
    • 2 years ago
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    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
    • 2 years ago
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    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
    • 2 years ago
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    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
    • 2 years ago
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    My suggestion would be to try polar although that has not given me any success so far.

  23. dumbcow
    • 2 years ago
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    we can assume the region is symmetric about x-axis correct ?

  24. vf321
    • 2 years ago
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    yes.

  25. vf321
    • 2 years ago
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    We know for a fact even

  26. dumbcow
    • 2 years ago
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    hmm ok im done :) last thought then is you have to break it up into different cases of possible regions

  27. vf321
    • 2 years ago
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    yes thats what I said you could do. But we'll see what SE says. I'll post the link soon.

  28. panlac01
    • 2 years ago
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    good job @dumbcow

  29. panlac01
    • 2 years ago
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    you sir, are the best

  30. vf321
    • 2 years ago
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    http://math.stackexchange.com/questions/187316/how-can-i-define-the-area-between-two-ellipses

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