How do you find the intersection of r = 1 - cos(theta)r=1−cos(θ), r^2 = 4cos(theta)r2=4cos(θ)?

1 Answer
A. S. Adikesavan
Nov 23, 2016

( 2 (sqrt2 - 1), cos^(-1) ( 3 - sqrt 2 ) ) = ( 0.8284, 80.12^o )(2(√2−1),cos−1(3−√2))=(0.8284,80.12o)

Explanation:

r > 0.

Eliminate r.

r^2=(1-cos theta )^2=4 cos thetar2=(1−cosθ)2=4cosθ

Solving,

cos theta=3-2sqrt2 to theta =arc cos (3-2sqrt2) to r =2((sqrt2-1)cosθ=3−2√2→θ=arccos(3−2√2)→r=2((√2−1).

So, the common point is

( 2 (sqrt2 - 1), cos^(-1) ( 3 - sqrt 2 ) ) = ( 0.8284, 80.12^o )(2(√2−1),cos−1(3−√2))=(0.8284,80.12o)

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