Question

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.

Answer 1

\[\frac{y^2}{y_1^2}+\frac{x^2}{x_1^2}=1\]\[\frac{y^2}{y_2^2}+\frac{(x-x_0)^2}{x_2^2}=1\]As stated in assumptions, \(x_0>x_1-x_2\), and the resulting inequalities may be expressed in terms of all subscripted constants.

Answer 2

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

Answer 3

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

Answer 4

@hero is here to save the day

Answer 5

count me out

Answer 6

x_0 is a point,
x_1 - x_2 is a distance

Answer 7

@Geometry_Hater

Answer 8

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

Answer 9

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

Answer 10

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

Answer 11

Yes that would get us the two points of intersection, I agree. I can do that. But that still doesn't define the area.

Answer 12

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

Answer 13

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

Answer 14

Yes. But say I have to elipses:

I admit that we can find a, b, and c. But how does that help us find the region in the middle?