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

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Answer 1 · 2016-10-27T07:29:27

We know the relations

$x=rcostheta and y =rsintheta$

Again $x^2+y^2=r^2$

where $(r,theta)$ are the polar coordinates of a point corresponding to the rectangular coordinates $(x,y)$

The given equation in rectanglar form is

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

$=>y^2=x^2-6xy+9y^2-x^2$

$=>8y^2=6xy$

$=>8r^2sin^2theta=6xxrcosthetaxxrsintheta$

$=>8sin^2theta=6costhetasintheta$

$=>4(1-cos2theta)=3sin2theta$

$=>3sin2theta+4cos2theta=4$

This is the polar form of the given equation.

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