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

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Answer 1 · 2017-06-04T20:20:11

The polar equation is $r = \frac{sin θ + 4 cos θ}{{sin}^{2} θ - 2 {cos}^{2} θ}$ $r=(sintheta+4costheta)/(sin^2theta-2cos^2theta)$

To convert from rectangular coordinates $(x, y)$ $(x,y)$ to polar coordinates $(r, θ)$ $(r,theta)$ , we use the following

$x = r cos θ$ $x=rcostheta$

$y = r sin θ$ $y=rsintheta$

Our equation is

$y = y^{2} - 2 x^{2} - 4 x$ $y=y^2-2x^2-4x$

$r sin θ = r^{2} {sin}^{2} θ - 2 r^{2} {cos}^{2} θ - 4 r cos θ$ $rsintheta=r^2sin^2theta-2r^2cos^2theta-4rcostheta$

Dividing by $r \neq 0$ $r!=0$

$sin θ = r {sin}^{2} θ - 2 r {cos}^{2} θ - 4 cos θ$ $sintheta=rsin^2theta-2rcos^2theta-4costheta$

$r ({sin}^{2} θ - 2 {cos}^{2} θ) = sin θ + 4 cos θ$ $r(sin^2theta-2cos^2theta)=sin theta+4costheta$

$r = \frac{sin θ + 4 cos θ}{{sin}^{2} θ - 2 {cos}^{2} θ}$ $r=(sintheta+4costheta)/(sin^2theta-2cos^2theta)$

How do you convert y=y^2-2x^2 -4x y=y2−2x2−4xy=y^2-2x^2 -4x into a polar equation?