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How do you convert $r = \frac{1}{8 - cos (θ)}$ $r = 1/(8 - cos(theta))$ into rectangular form?

1 Answer

Dec 7, 2015

$8 \sqrt{x^{2} + y^{2}} - x = 1$ $8sqrt(x^2+y^2)-x=1$

Explanation:

In converting between polar and rectangular forms:
$XXX r = \sqrt{x^{2} + y^{2}}$ $color(white)("XXX")r=sqrt(x^2+y^2)$
and
$XXX cos (θ) = \frac{x}{r} = \frac{x}{\sqrt{x^{2} + y^{2}}}$ $color(white)("XXX")cos(theta)=x/r = x/(sqrt(x^2+y^2))$

Therefore, given
$XXX r = \frac{1}{8 - cos (θ)}$ $color(white)("XXX")r = 1/(8-cos(theta))$

We have
$XXX r = \frac{1}{8 - \frac{x}{r}}$ $color(white)("XXX")r = 1/(8-x/r)$

$XXX r = \frac{1}{\frac{8 r - x}{r}}$ $color(white)("XXX")r= 1/((8r-x)/r)$

$XXX r = \frac{r}{8 r - x}$ $color(white)("XXX")r= r/(8r-x)$

$XXX 8 r - x = 1$ $color(white)("XXX")8r-x=1$

$XXX 8 (\sqrt{x^{2} + y^{2}}) - x = 1$ $color(white)("XXX")8(sqrt(x^2+y^2))-x = 1$

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