What the is the polar form of $y^2 = (x-1)^2/y-x^2$ ?

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Answer 1 · 2017-01-05T08:04:41

$r^3sintheta-r^2cos^2theta+2rcostheta-1=0$

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

$x=rcostheta$ , $y=rsintheta$ and $r^2=x^2+y^2$ .

Hence, $y^2=((x-1)^2)/y-x^2$ can be written as

$(rsintheta)^2=((rcostheta-1)^2)/(rsintheta)-(rcostheta)^2$

or $r^2sin^2theta=((r^2cos^2theta-2rcostheta+1))/(rsintheta)-r^2cos^2theta$

or $r^2sin^2theta+r^2cos^2theta=((r^2cos^2theta-2rcostheta+1))/(rsintheta)$

or $r^2=((r^2cos^2theta-2rcostheta+1))/(rsintheta)$

or $r^3sintheta-r^2cos^2theta+2rcostheta-1=0$

What the is the polar form of y^2 = (x-1)^2/y-x^2 y^2 = (x-1)^2/y-x^2 ?