How do you convert $2 y = 7 x^{2} - 2 x$ $2y= 7x^2-2x$ into a polar equation?

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Answer 1 · 2016-12-01T17:16:08

$r = \frac{2 (sin θ + cos θ)}{7 {cos}^{2} θ}$ $r=(2(sintheta+costheta))/(7cos^2theta)$

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

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

Hence $2 y = 7 x^{2} - 2 x$ $2y=7x^2-2x$ van be written as

$2 r sin θ = 7 {(r cos θ)}^{2} - 2 r cos θ$ $2rsintheta=7(rcostheta)^2-2rcostheta$

or $2 r sin θ + 2 r cos θ = 7 r^{2} {cos}^{2} θ$ $2rsintheta+2rcostheta=7r^2cos^2theta$

or $2 r (sin θ + cos θ) = 7 r^{2} {cos}^{2} θ$ $2r(sintheta+costheta)=7r^2cos^2theta$

or $r = \frac{2 (sin θ + cos θ)}{7 {cos}^{2} θ}$ $r=(2(sintheta+costheta))/(7cos^2theta)$

How do you convert 2y= 7x^2-2x 2y=7x2−2x2y= 7x^2-2x into a polar equation?