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

1 Answer
Jun 24, 2018

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)