How do you convert #r=25/( 5-3cos(theta-30))# into rectangular form?

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Answer 1 · 2016-10-26T10:42:29

#5sqrt(x^2+y^2)-(3sqrt3)/2x -1/2y=25#

Polar coordinates #(r,theta)# and Cartesian coordinates #(x,y)# are related as

#x=rcostheta# and #y=rsintheta#, i.e. #r^2=x^2+y^2# and #tantheta=y/x#

Hence, #r=25/(5-3cos(theta-30^o))# can be written as

#r(5-3cos(theta-30^o))=25#

or #r(5-3(costhetacos30^o +sinthetasin30^o))=25#

or #r(5-3(costhetaxxsqrt3/2 +sinthetaxx1/2))=25#

or #5r-(3sqrt3)/2rcostheta -1/2rsintheta=25#

or #5sqrt(x^2+y^2)-(3sqrt3)/2x -1/2y=25#