What the is the polar form of $y = 1/y^3-xy+x^2/y$ ?

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Answer 1 · 2018-02-18T11:48:23

Polar form is $r^4sin^4theta(1+rcos theta)-r^2cos^2theta*sin^2theta =1$

$y=1/y^3-xy+x^2/y$ Multiplying by $y^3$ on both sides we get,

$y^4=1-xy^4+x^2*y^2$ or

$y^4+xy^4=1+x^2*y^2$ or

$y^4(1+x)=1+x^2*y^2$ or

$y^4(1+x)-x^2*y^2 =1$

Polar form: $x = r costheta and y = r sin theta and x^2+y^2=r^2$

$r^4sin^4theta(1+rcos theta)-r^4cos^2theta*sin^2theta =1$

Polar form is

$r^4sin^4theta(1+rcos theta)-r^2cos^2theta*sin^2theta =1$ [Ans]

What the is the polar form of y = 1/y^3-xy+x^2/y y = 1/y^3-xy+x^2/y ?