How do you convert $r = \frac{10}{4 - 7 cos (θ)}$ $r = 10/(4 - 7cos(theta))$ to rectangular form?

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Answer 1 · 2016-05-06T18:04:35

$33 x^{2} - 16 y^{2} + 140 x + 100 = 0$ $33x^2 -16y^2+140x+100=0$

rectangular and polar form are related as follows:

$x = r cos θ$ $x= r cos theta$ and $y = r sin θ$ $y= rsin theta$ , so $r^{2} = x^{2} + y^{2}$ $r^2= x^2 +y^2$

Accordingly, the given expression can be reworked as follows:

$4 r - 7 r cos θ = 10$ $4r -7rcos theta=10$

$4 r = 10 + 7 x$ $4r= 10+7x$ Square both sides,

$16 r^{2} = {(10 + 7 x)}^{2}$ $16r^2= (10+7x)^2$

$16 x^{2} + 16 y^{2} = 100 + 140 x + 49 x^{2}$ $16x^2+16y^2= 100+140x +49x^2$

$33 x^{2} - 16 y^{2} + 140 x + 100 = 0$ $33x^2 -16y^2+140x+100=0$

How do you convert r = 10/(4 - 7cos(theta))r=104−7cos(θ)r = 10/(4 - 7cos(theta)) to rectangular form?