anonymous
  • anonymous
Area of triangle formed by x^2 -y^2 +2x = 1 and x+y=3??
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
that first equation is that of a hyperbola, the second equation is a line.... how do they form a triangle? their intersection(s)?
anonymous
  • anonymous
Exactly...my point.
anonymous
  • anonymous
No ...wait. It ain't a hyperbola.

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anonymous
  • anonymous
|dw:1332929032807:dw| this hyperbola has asympotes with slopes plus/minus 1. the slope of the line is negative 1. it is parallel to one of the asymptotes....
anonymous
  • anonymous
at most the intersection is at 1 point....
dumbcow
  • dumbcow
http://www.wolframalpha.com/input/?i=plot+sqrt%28x^2%2B2x-1%29%2C-sqrt%28x^2%2B2x-1%29%2C+3-x
dumbcow
  • dumbcow
what triangle are you referring to?
anonymous
  • anonymous
Oh! Sorry! I just saw the question again...its: \[x^2 - y^2 + 2y = 1\]
dumbcow
  • dumbcow
i think you would have to integrate to find are under both curves they intersect at x = 5/4 \[A = \int\limits_{\sqrt{2}-1}^{5/4}\sqrt{x^{2}+2x-1}dx + \int\limits_{5/4}^{3} (3-x)dx\]
anonymous
  • anonymous
now that new equation is still a hyperbola but a degenerate hyperbola. the line is still parallel to the asymptotes...
anonymous
  • anonymous
How is still a hyperbola? We can write it like this, right? |dw:1332930016219:dw|
dumbcow
  • dumbcow
no you can't write it like that...its a hyperbola dpalnc described it above i have shown a graph , its def a hyperbola
anonymous
  • anonymous
that's a degenerate hyperbola.
anonymous
  • anonymous
But degenerate hyperbola are two lines...right?
anonymous
  • anonymous
wait. a degenerate hyperbola is basically the two intersecting "asymptotes". so the triangle formed must be the two points intersected by the third line and the third point is the intersection of the "asymptotes"....
anonymous
  • anonymous
i guess the question now is where do the intersecting "asymptotes" intersect? so you'll get your limits
anonymous
  • anonymous
But when I am taking out the asymptotes, one line is coming parallel to x + y= 3. How do I solve now?
anonymous
  • anonymous
|dw:1332931349816:dw|... you're right, the line only intersect the "hyperbola" at 1 point.
anonymous
  • anonymous
but the center of the hyperbola is (0,1) this is the where the asymptotes intersect. you at least have 2 points. where's the third vertice? sorry man, this is too much math for me.
anonymous
  • anonymous
No problem...thanks for your help anyway! :D
anonymous
  • anonymous
could the triangle be the triangle created by the hyperbola, x+y=3, and the x-axis?
anonymous
  • anonymous
Do you know the answer?

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