How do you graph r(2 + cos theta) = 1r(2+cosθ)=1?

1 Answer
A. S. Adikesavan
Mar 31, 2016

This is polar equation (1/2)/r=1+(1/2)cos theta12r=1+(12)cosθ of an ellipse referred to a focus as pole and major axis (in the direction of nearer end) as initial line. Major axis = 4/3 and eccentricity = 1/2..

Explanation:

Compare with the standard form l/r=1+e cos thetalr=1+ecosθ, the rearranged equation . (1/2)/r=1+(1/2)cos theta12r=1+(12)cosθ

e=1/2 and l = a (1-e^2)=1/2e=12andl=a(1−e2)=12.
a=1/(2(1-1/4))=2/3a=12(1−14)=23.
O(0, 0) is a focus S.
The other focus S'(2ae, pi) = (2/3, pi)(2ae,π)=(23,π)
The center C is the midpoint (1/3, pi)(13,π), in-between.
Put theta =0, piθ=0,π in the equation to get ends of the major axis,
A(1/3, 0), A'(1, pi)

Put r = a =2/3 to get the ends of the minor axis, B(2/3, 2pi/3), B'(2/3, -2pi/3)#.

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