How do you convert $y = - x^{2} + 3 x y - 4$ $y= -x^2+3xy-4$ into a polar equation?

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Answer 1 · 2016-12-29T15:24:21

$r^{2} cos θ (3 sin θ - 1) + r sin θ + 4 = 0$ $r^2costheta(3sintheta-1) +rsintheta+4=0$

Substitute:

$x = r cos θ$ $x=rcostheta$
$y = r sin θ$ $y=rsintheta$

$r sin θ = - r^{2} cos θ + 3 r^{2} sin θ cos θ - 4$ $rsintheta = -r^2costheta +3r^2sinthetacostheta-4$

or:

$r^{2} cos θ (3 sin θ - 1) + r sin θ + 4 = 0$ $r^2costheta(3sintheta-1) +rsintheta+4=0$

How do you convert y= -x^2+3xy-4 y=−x2+3xy−4y= -x^2+3xy-4 into a polar equation?