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

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Answer 1 · 2016-04-15T20:23:14

$r^{2} {cos}^{2} θ - r^{3} {sin}^{3} θ - 6 r cos θ + 9 = 0$ $r^2cos^2theta-r^3sin^3theta-6rcostheta+9=0$

$y^{3} = {(x - 3)}^{2}$ $y^3=(x-3)^2$

$y^{3} = x^{2} - 6 x + 9$ $y^3=x^2-6x+9$

$0 = x^{2} - y^{3} - 6 x + 9$ $0=x^2-y^3-6x+9$

Use the formulas:
$x = r cos θ and y = r sin θ$ $x=rcostheta and y=rsin theta$

$0 = r^{2} {cos}^{2} θ - r^{3} {sin}^{3} θ - 6 r cos θ + 9$ $0=r^2cos^2theta-r^3sin^3theta-6rcostheta+9$

What the is the polar form of y^2 = (x-3)^2/y y2=(x−3)2yy^2 = (x-3)^2/y ?