What the is the polar form of $y^2 = (x-3)^2/y-x^2$ ?

Question

1483 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-04-22T01:02:26

$r^3sin^3theta-r^2cos^2theta-9+6rcostheta+r^3cos^2thetasintheta=0$

$y^2=((x-3)^2)/y - x^2$

Put $x=rcostheta$ and $y=rsintheta$ ; we get :-

$r^2 sin^2theta=((rcostheta-3)^2)/(rsintheta) -r^2cos^2theta$

$rArrr^3sin^3theta-r^2cos^2theta-9+6rcostheta+r^3cos^2thetasintheta=0$

What the is the polar form of y^2 = (x-3)^2/y-x^2 y^2 = (x-3)^2/y-x^2 ?