How do you change the polar equation $r (2 + cos θ) = 1$ $r(2+costheta)=1$ into rectangular form?

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Answer 1 · 2016-10-31T22:37:27

${(x \sqrt{3} + \frac{1}{\sqrt{3}})}^{2} + 4 y^{2} - \frac{4}{3} = 0$ $(xsqrt(3)+1/sqrt(3))^2+4y^2-4/3=0$

Giving the pass equations

${(x=rcostheta),(y=rsintheta):}$

$r(2+costheta)=1 = r(2+x/r)=2r+x=1$ then

$r = (1-x)/2->x^2+y^2=(1-2x+x^2)/4$ and finally

$3x^2+4y^2+2x-1=0$ or

$(xsqrt(3)+1/sqrt(3))^2+4y^2-4/3=0$ which is an ellipse.

How do you change the polar equation r(2+costheta)=1r(2+cosθ)=1r(2+costheta)=1 into rectangular form?