How do you convert $2 y = - 3 x^{2} + 9 x$ $2y= -3x^2+9x$ into a polar equation?

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Answer 1 · 2017-09-19T09:17:59

$r = 3 sec θ - \frac{2}{3} sec θ tan θ$ $r=3sectheta-2/3secthetatantheta$

The relation between polar coordinates $(r, θ)$ $(r,theta)$ and Carrtesian coordinates $(x, y)$ $(x,y)$ is given by

$x = r cos θ$ $x=rcostheta$ , $y = r sin θ$ $y=rsintheta$ and $x^{2} + y^{2} = r^{2}$ $x^2+y^2=r^2$

Hence, $2 y = - 3 x^{2} + 9 x$ $2y=-3x^2+9x$ can be written as

$2 r sin θ = - 3 r^{2} {cos}^{2} θ + 9 r cos θ$ $2rsintheta=-3r^2cos^2theta+9rcostheta$

or $2 sin θ = - 3 r {cos}^{2} θ + 9 cos θ$ $2sintheta=-3rcos^2theta+9costheta$

or $3 r {cos}^{2} θ = 9 cos θ - 2 sin θ$ $3rcos^2theta=9costheta-2sintheta$

or $r = \frac{9 cos θ}{3 {cos}^{2} θ} - \frac{2 sin θ}{3 {cos}^{2} θ}$ $r=(9costheta)/(3cos^2theta)-(2sintheta)/(3cos^2theta)$

or $r = 3 sec θ - \frac{2}{3} sec θ tan θ$ $r=3sectheta-2/3secthetatantheta$

How do you convert 2y= -3x^2+9x 2y=−3x2+9x2y= -3x^2+9x into a polar equation?