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

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Answer 1 · 2016-12-06T06:11:20

$r = - \frac{3}{2} \frac{tan θ sec θ}{1 - 2 tan θ}$ $r = -3/2(tan theta sec theta)/(1-2 tan theta)$ , representing a hyperbola. Graph is inserted.

The conversion formula is (x, y) = r (cos theta, sin theta)#

After conversion,

$- 3 sin θ = 2 r cos θ (cos θ + 2 sin θ)$ $-3sin theta = 2rcos theta (cos theta + 2 sin theta)$ . Explicitly,

$r = - \frac{3}{2} \frac{tan θ}{cos θ + 2 sin θ}$ $r=-3/2tan theta/(cos theta+2 sin theta$

$= = - \frac{3}{2} \frac{tan θ sec θ}{1 - 2 tan θ}$ $== -3/2(tan theta sec theta)/(1-2 tan theta$ .
graph{3y+2x^2+4xy=0 [-5.53, 5.53, -2.765, 2.765]}

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