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Les Paul
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#1
Report Thread starter 14 years ago
#1
I'd appreciate if someone could help me do this question.


Show that the 2 tangents to the curve
3y^2 = x^2(2x+1)
at the origin intersect at an angle of 3/pi


Thanks a lot
Chris
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RichE
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#2
Report 14 years ago
#2
(Original post by Les Paul)
I'd appreciate if someone could help me do this question.


Show that the 2 tangents to the curve
3y^2 = x^2(2x+1)
at the origin intersect at an angle of 3/pi


Thanks a lot
Chris
The equation of the top half of the graph is

rt(3) y = x rt(2x+1)

Differentiating we get

rt(3) dy/dx = rt(2x+1) + x/rt(2x+1)

so that dy/dx = 1/rt(3)

This is the tangent of the angle the curve makes with the x-axis.

So the top part of the curve makes arctan(1/rt(3)) = pi/6 with the x-axis

The angle between that tangent and the other tangent (which is its reflection in the x-axis) is twice this - namely pi/3
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Les Paul
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#3
Report Thread starter 14 years ago
#3
Ahhh thanks very much
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