What the is the polar form of $y = \frac{y}{x} + {(x - 2 y)}^{2}$ $y = y/x+(x-2y)^2$ ?

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What the is the polar form of $y = \frac{y}{x} + {(x - 2 y)}^{2}$ $y = y/x+(x-2y)^2$ ?

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Answer 1 · 2018-03-10T20:09:32

We know that $x = r cos θ$ $x= rcostheta$ and $y = r sin θ$ $y = rsintheta$ .

$r sin θ = \frac{r sin θ}{r cos θ} + {(r cos θ - 2 r sin θ)}^{2}$ $rsintheta = (rsintheta)/(rcostheta) + (rcostheta - 2rsintheta)^2$

$r sin θ = tan θ + r^{2} {cos}^{2} θ - 4 r^{2} sin θ cos θ + 4 r^{2} {sin}^{2} θ$ $rsintheta = tan theta + r^2cos^2theta - 4r^2sinthetacostheta + 4r^2sin^2theta$

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What the is the polar form of $y = \frac{y}{x} + {(x - 2 y)}^{2}$ $y = y/x+(x-2y)^2$ ?

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What the is the polar form of $y = \frac{y}{x} + {(x - 2 y)}^{2}$ $y = y/x+(x-2y)^2$ ?