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

Question

1068 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-01-05T08:04:41

#r^3sintheta-r^2cos^2theta+2rcostheta-1=0#

The relation between polar coordinates #(r,theta)# and corresponding Cartesian coordinates #(x,y)# is given by

#x=rcostheta#, #y=rsintheta# and #r^2=x^2+y^2#.

Hence, #y^2=((x-1)^2)/y-x^2# can be written as

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

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

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

or #r^2=((r^2cos^2theta-2rcostheta+1))/(rsintheta)#

or #r^3sintheta-r^2cos^2theta+2rcostheta-1=0#