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 · 2016-11-08T10:56:10

The cartesian form is $3 x^{2} - 12 x + 4 y^{2} - 36 = 0$ $3x^2-12x+4y^2-36=0$

Explanation:

The formulae used are $r^{2} = x^{2} + y^{2}$ $r^2=x^2+y^2$
and $x = r cos θ$ $x=rcostheta$ and $y = r sin θ$ $y=rsintheta$

Rewring the polar form as $r (2 - cos θ) = 6$ $r(2-costheta)=6$
Replacing $cos θ = \frac{x}{r}$ $costheta=x/r$
We get $r (2 - \frac{x}{r}) = 6$ $r(2-x/r)=6$ $\Rightarrow$ $=>$ $2 r - x = 6$ $2r-x=6$
$\Rightarrow$ $=>$ $2 r = x + 6$ $2r=x+6$
replacing $r$ $r$
$2 \sqrt{x^{2} + y^{2}} = x + 6$ $2sqrt(x^2+y^2)=x+6$
Squaring both sides
$4 (x^{2} + y^{2}) = {(x + 6)}^{2}$ $4(x^2+y^2)=(x+6)^2$
$4 x^{2} + 4 y^{2} = x^{2} + 12 x + 36$ $4x^2+4y^2=x^2+12x+36$
$3 x^{2} - 12 x + 4 y^{2} - 36 = 0$ $3x^2-12x+4y^2-36=0$
graph{3x^2-12x+4y^2-36=0 [-7.9, 7.9, -3.95, 3.95]}

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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?