How do you convert $R = \frac{6}{2 + cos (θ)}$ $R=6/(2+cos(theta))$ into cartesian form?

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How do you convert $R = \frac{6}{2 + cos (θ)}$ $R=6/(2+cos(theta))$ into cartesian form?

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Answer 1 · 2017-11-28T12:55:13

$3 x^{2} + 12 x + 4 y^{2} - 36 = 0$ $3x^2+12x+4y^2-36=0$

Explanation:

$r = \frac{6}{2 + cos (θ)}$ $r=6/(2+cos(theta))$

$2 r + r cos (θ) = 6$ $2r+rcos(theta)=6$

For polar coordinates; $r$ $r$ is equal to $\sqrt{x^{2} + y^{2}}$ $sqrt(x^2+y^2)$ and $r cos (θ)$ $rcos(theta)$ is equal to $x$ $x$ in cartesian ones. Hence,

$2 \sqrt{x^{2} + y^{2}} + x = 6$ $2sqrt(x^2+y^2)+x=6$

$2 \sqrt{x^{2} + y^{2}} = 6 - x$ $2sqrt(x^2+y^2)=6-x$

$4 \cdot (x^{2} + y^{2}) = {(6 - x)}^{2}$ $4*(x^2+y^2)=(6-x)^2$

$4 x^{2} + 4 y^{2} = x^{2} - 12 x + 36$ $4x^2+4y^2=x^2-12x+36$

$3 x^{2} + 12 x + 4 y^{2} - 36 = 0$ $3x^2+12x+4y^2-36=0$

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How do you convert $R = \frac{6}{2 + cos (θ)}$ $R=6/(2+cos(theta))$ into cartesian form?

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Explanation:

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How do you convert $R = \frac{6}{2 + cos (θ)}$ $R=6/(2+cos(theta))$ into cartesian form?