How do you convert # r(2 - cos theta) = 2# into a rectangular equation?

1 Answer
A. S. Adikesavan
Apr 20, 2016

#3 x^2+4 y^2-4 x -4 = 0#

Explanation:

#2 r = r cos theta + 2=x+2#.

So, #4 r^2=(x+2)^2#.
#4 (x^2+y^2)=x^2+4x+4#
#3 x^2+4 y^2-4 x -4 = 0#

The given equation is #1/r=1-(1/2) cos theta#
This represents an ellipse with eccentricity #e = 1/2# and semi-latus rectum l = 1. Semi-major axis a = #l/(1-e^2)=4/3#.
The negative sign indicates that the initial line is reversed, from pole (a focus) to the center of the ellipse.

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