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an ellipse of eccentricity [2sqrt(2)]/3 is inscribed in a circle and a point within the circle is chosen at random.what is the probability that the point lies outside the ellipse
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do you recall what the eccentricity measures?
e = c/a .... focal length in terms of vertex length
|dw:1400510949235:dw|
yeah its\[\sqrt{(1-\frac{ b ^{2} }{ a ^{2}}}\]
the area of the circle if pi a^2
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yeah if a=b then its pia^2 then?
if a=b its a circle, and e=1
the circle that the ellipse is drawn inside of has a radius of a
yeah ok
|dw:1400511115595:dw|
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P(inside ellipse) = area ellipse/ area circle so 1- that is outside the ellipse
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