How do you convert $x y = x - 2 y + y^{2}$ $xy=x-2y+y^2$ into a polar equation?

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Answer 1 · 2016-02-20T06:54:38

$r = \frac{cos θ - 2 \cdot sin θ}{sin θ \cdot cos θ - {sin}^{2} θ}$ $r=(cos theta-2*sin theta)/(sin theta* cos theta-sin^2 theta)$

just use the transformation equivalent

$x = r cos θ$ $x=r cos theta$ and $y = r sin θ$ $y=r sin theta$ in the equation

$x y = x - 2 y + y^{2}$ $xy=x-2y+y^2$

so that $x y = x - 2 y + y^{2}$ $xy=x-2y+y^2$ becomes

$r cos θ \cdot r sin θ = r cos θ - 2 \cdot r sin θ + {(r sin θ)}^{2}$ $r cos theta* r sin theta=r cos theta-2* r sin theta+ (r sin theta)^2$

canceling some of the r

$r \cdot cos θ \cdot sin θ = cos θ - 2 \cdot sin θ + r {sin}^{2} θ$ $r* cos theta* sin theta= cos theta-2* sin theta+ r sin^2 theta$

simplify to obtain

$r \cdot (cos θ \cdot sin θ - {sin}^{2} θ) = cos θ - 2 \cdot sin θ$ $r* (cos theta* sin theta- sin^2 theta)= cos theta-2* sin theta$

and

$r = \frac{cos θ - 2 \cdot sin θ}{cos θ \cdot sin θ - {sin}^{2} θ}$ $r=(cos theta-2* sin theta)/(cos theta* sin theta- sin^2 theta)$

have a nice day!

How do you convert xy=x-2y+y^2 xy=x−2y+y2xy=x-2y+y^2 into a polar equation?