How do you convert $5 y = - 3 x^{2} - 2 x$ $5y= -3x^2-2x$ into a polar equation?

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Answer 1 · 2017-03-06T10:00:26

$r = - \frac{1}{3} sec θ (5 tan θ - 2)$ $r=-1/3sectheta(5tantheta-2)$

for Cartesian to /from Polar form we use teh eqns.

$r^{2} = x^{2} + y^{2}$ $r^2=x^2+y^2$

$x = r cos θ$ $x=rcostheta$

$y = r sin θ$ $y=rsintheta$

$5 y = - 3 x^{2} - 2 x$ $5y=-3x^2-2x$

becomes;
$5 r sin θ = - 3 {(r cos θ)}^{2} - 2 r cos θ$ $5rsintheta=-3(rcostheta)^2-2rcostheta$

$5cancel(r)sintheta=-3cancel(r^2)^rcos^2theta-2cancel(r)costheta$

$5sintheta=-3rcos^2theta-2costheta$

$3rcos^2theta=-5sintheta-2costheta$

$r=(-5sintheta-2costheta)/(3cos^2theta)$

$r=(-5sintheta)/(3cos^2theta) -(2costheta)/(3cos^2theta)$

$r=-5/3tanthetasectheta-2/3sectheta$

$r=-1/3sectheta(5tantheta-2)$

How do you convert 5y= -3x^2-2x 5y=−3x2−2x5y= -3x^2-2x into a polar equation?