How do you convert $- x y = x + 4 y^{2}$ $-xy=x+4y^2$ into a polar equation?

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Answer 1 · 2016-04-15T18:28:48

$4 r^{2} {sin}^{2} θ + r^{2} cos θ sin θ + r cos θ = 0$ $4r^2sin^2theta+r^2costhetasintheta+rcostheta=0$

$- x y = x + 4 y^{2}$ $-xy=x+4y^2$

Use equations $x = r cos θ, y = r sin θ$ $x=rcos theta,y=rsin theta$

$- r cos θ \times r sin θ = r cos θ + 4 r^{2} {sin}^{2} θ$ $-rcos theta xx rsintheta=rcos theta+4r^2sin^2theta$

$- r^{2} cos θ sin θ = r cos θ + 4 r^{2} {sin}^{2} θ$ $-r^2costhetasintheta=rcostheta+4r^2sin^2theta$

$0 = 4 r^{2} {sin}^{2} θ + r^{2} cos θ sin θ + r cos θ$ $0=4r^2sin^2theta+r^2costhetasintheta+rcostheta$

How do you convert -xy=x+4y^2 −xy=x+4y2-xy=x+4y^2 into a polar equation?