How do you convert $(x^2 + y^2)^(3/2) = 3y^2 - 3x^2$ to polar form?

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Answer 1 · 2016-02-06T14:53:57

$r^3= 3 cos 2theta$

In polar form x equals $r cos theta$ and y equals $rsin theta$ . Accordingly, $x^2+y^2= r^2$ and $y^2-x^2= cos^2 theta -sin^2 theta = cos 2theta$

The given expression in polar form would thus be $r^3= 3 cos 2theta$

How do you convert (x^2 + y^2)^(3/2) = 3y^2 - 3x^2(x^2 + y^2)^(3/2) = 3y^2 - 3x^2 to polar form?