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What the is the polar form of #y^2 = (x-1)^2/y-3x^2y #?

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Jun 24, 2018

Answer:

#r^2=1/sin^3theta(costheta(rcostheta(1+3r(-1+cos^2theta)-2))+1/r)#

any of the simplifications are answers

Explanation:

using the formula

#x=rcostheta#

and

#y=rsintheta#

substitute

#(rsintheta)^2=((rcostheta-1)^2)/(rsintheta)-3(rcostheta)^2rsintheta#

expand brackets and simplify

#r^2sin^2theta=(r^2cos^2theta-2rcostheta+1)/(rsintheta)-3r^2cos^2thetarsintheta*(rsintheta)/(rsintheta)#

#r^2sin^2theta=(r^2cos^2theta-2rcostheta+1)/(rsintheta)-(3r^2cos^2thetarsin^2theta)/(rsintheta)#

#r^2sin^2theta=(r^2cos^2theta-2rcostheta+1-3r^2cos^2thetarsin^2theta)/(rsintheta)#

simplify

#r^2sin^2theta=(r^2cos^2theta-2rcostheta+1-3r^3cos^2thetasin^2theta)/(rsintheta)#

factorise

#r^2sin^2theta=(r(rcos^2theta-2costheta-3r^2cos^2thetasin^2theta)+1)/(rsintheta)#

#r^2sin^2theta=(cancel(r)(rcos^2theta-2costheta-3r^2cos^2thetasin^2theta))/(cancel(r)sintheta)+(1)/(rsintheta)#

multiply by #1/sin^2theta#

#1/cancel(sin^2theta)*r^2cancel(sin^2theta)=1/sin^2theta((rcos^2theta-2costheta-3r^2cos^2thetasin^2theta)/(sintheta)+(1)/(rsintheta))#

expand the brackets

#r^2=(rcos^2theta-2costheta-3r^2cos^2thetasin^2theta)/(sin^3theta)+(1)/(rsin^3theta)#

factorise

#r^2=1/sin^3theta(rcos^2theta-2costheta-3r^2cos^2thetasin^2theta+1/r)#

convert #sin^2theta# to #1-cos^2theta#

#r^2=1/sin^3theta(rcos^2theta-2costheta-3r^2cos^2theta(1-cos^2theta)+1/r)#

expand the #-3r^2cos^2theta(1-cos^2theta)# brackets

#r^2=1/sin^3theta(rcos^2theta-2costheta-3r^2cos^2theta+3r^2cos^4theta+1/r)#

factorise the terms with #rcos^2theta# in it out

#r^2=1/sin^3theta(rcos^2theta(1-3r+3rcos^2theta)-2costheta+1/r)#

factorise

#r^2=1/sin^3theta(costheta(rcostheta(1+3r(-1+cos^2theta)-2))+1/r)#

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