How do you convert #5y= x -2xy # into a polar equation?

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Answer 1 · 2018-05-13T11:32:41

#r=(costheta-5sintheta)/(sin(2theta))#

For this we will use the two equations:
#x=rcostheta#
#,y=rsintheta#

#5rsintheta=rcostheta-2(rcos theta)(rsintheta)#

#5rsintheta=rcostheta-2r^2costhetasintheta#

#5sintheta=costheta-2rcosthetasintheta#

#2rcosthetasintheta=costheta-5sintheta#

#r=(costheta-5sintheta)/(2costhetasintheta)#

#r=(costheta-5sintheta)/(sin(2theta))#