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Polar coordinates: r = 1/(1+2costheta)

Working through this equation, I'm happy with identifying asymptotes and points at theta = 0, Pi/2, Pi, 3/2Pi.
So it crosses x at 1/3 and 1 and y at 1 and -1 with asymptotes theta =2/3 Pi and 4/3 Pi (where cos =-1/2)
Now when you plot it you get symmetry half way between x =1/3 and 1 and you get asymptotes that are parallel to those from the origin, but centred at the line of symmetry.

I have a disconnect as to how I can justify the asymptotes running through x=2/3 as opposed to being at zero, since all the maths works from the origin. But the positive r line goes from y=-1 to x=+1/3 to y=1 so crosses the asymptotes as defined from the origin.

Reply 1

mqb2766

Original post by Phys&Math_Ethan

Cant say its the most efficient reason, but transfering back to cartesian you get
x^2+y^2=(1-2x)^2
and rearranging gives the shifted hyperbola

Another way would be to note the polar form of
ax+by-c=0
would be
r = c/(bsin(theta)+acos(theta))
Asymptotically as theta->2pi/3, then b=1/sin(2pi/3)=2/sqrt(3). Similarly c=4/3 to give the line
2x+2/sqrt(3)y=4/3
which gives the xaxis crossing point of the asymptote