How do you convert $3y= 3x^2-6xy+3x$ into a polar equation?

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Answer 1 · 2016-06-17T12:44:05

$=>r=(tantheta-1)/(costheta-2sintheta)$

We know that if a point in X-Y plane has rectangular coordinate
$(x,y)$ and its polar coordinate is $(r,theta)$ ,then we have the relation

$x=rcostheta and y =rsintheta$

and $r=sqrt(x^2+y^2)$

Now the given equation is

$3y= 3x^2-6xy+3x$

Dividing bothsides by 3

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

Inserting $x=rcostheta and y =rsintheta$ in the given equation

$rsintheta=r^2cos^2theta-2rcostheta*rsintheta+rcostheta$

Dividing both sides by $rcostheta$

$=>tantheta=rcostheta-2rsintheta+1$

$=>r=(tantheta-1)/(costheta-2sintheta)$

How do you convert 3y= 3x^2-6xy+3x 3y= 3x^2-6xy+3x into a polar equation?