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

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Answer 1 · 2016-12-05T05:06:37

#r=sec theta( (3tan theta+1)/(3-6tan theta))#. The graph of the given hyperbola is inserted.

The conversion formula is

#(x, y)=r(cos theta, sin theta)#,

Rearranging, the given equation becomes

#x= r cos theta)=(3y-x)?(3x-6y)#

#=(3 sin theta - cos theta)/(3cos theta-6 sin theta)#

#=(3 tan theta+1)(3-6 tan theta)#, So, explicitly,

#r=sec theta( (3tan theta+1)/(3-6tan theta))#

graph{3x^2-6xy-3y-x=0 [-10, 10, -5, 5]}