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Answer 1 · 2017-12-11T13:26:28

#y^2=2x+1#

#r*(4-4cos(theta))=4#

#4r-4rcos(theta)=4#

After using #x=rcos(theta)# and #r=sqrt(x^2+y^2)# substitution,

#4sqrt(x^2+y^2)-4x=4#

#4sqrt(x^2+y^2)=4x+4#

#sqrt(x^2+y^2)=x+1#

#x^2+y^2=(x+1)^2#

#x^2+y^2=x^2+2x+1#

#y^2=2x+1#