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easy math?
Find the point of intersection of the curve xy=1, 2x^2-y=1, and determine the angles of which they cut each other.
the given answer is
(1,1),(-1,-1)
angle of intn.=arc tan 3=71°30'
1 Answer
- ?Lv 64 years ago
y = 2x^2 - 1
x(2x^2-1) = 1
2x^3 - x = 1
(1,1) works, but I do not see how (-1,-1) does.
1 - dy/dx = -1/x^2
2 - dy/dx = 4x
at x = 1
tan (theta) = (4-(-1)) / (1 + 4 * -1)
tan (theta) = 5/(-3)
theta = arc tan (-5/3)
theta = -59.04 degrees or (+180) = 120.96 deg = 120d 58'
but they probably want the acute angle so (see Pope's comment) 59d 2'
tan = (m1-m2)/(1+m1m2) when m1>m2
tan = (m2-m1)/(1+m1m2) when m2>m1
Again, I am not sure how they got their answer.
Source(s): I could easily be wrong
