What the is the polar form of $y = y/x-(x+y)^2$ ?

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Answer 1 · 2016-01-23T09:05:31

$r^2(1+sin 2theta)+r sin theta-tan theta=0$

To convert, use $y=r sin theta$ and $x=r cos theta$

replace all $x$ and $y$ the equivalents

$y=y/x-(x+y)^2$
$r sin theta=(r sin theta)/(r cos theta)-( r cos theta+ r sin theta)^2$

$r sin theta=(cancelr sin theta)/(cancelr cos theta)-( r^2)( cos theta+ sin theta)^2$

$r sin theta=(sin theta)/( cos theta)-( r^2)( cos^2 theta+ sin^2 theta+2 sin theta cos theta)$

$r sin theta= tan theta-r^2(1+sin 2theta)$

$r^2(1+sin 2theta)+r sin theta-tan theta=0$

What the is the polar form of y = y/x-(x+y)^2 y = y/x-(x+y)^2 ?