How do you convert #4y+2=(3x+4)^2+(2y-1)^2# into polar form?

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Answer 1 · 2018-04-15T02:27:03

See below

The conversion from Rectangular to Polar:
#x=rcostheta#
#y=rsintheta#

Substitute for #x# and #y# and solve for #r#:
#4(rsintheta)+2= (3(rcostheta)+4)^2+(2(rsintheta)-1)^2#

#4rsintheta+2= (3rcostheta+4)^2+(2rsintheta-1)^2#

#4rsintheta+2= 9r^2cos^2theta+24rcostheta+16+4r^2sin^2theta-4rsintheta+1#

#9r^2cos^2theta+24rcostheta+4r^2sin^2theta-8rsintheta+15=0#

#r(9rcos^2theta+24costheta+4rsin^2theta-8sintheta)=-15#

At this point either #r# is equal to #-15# or #(9rcos^2theta+24costheta+4rsin^2theta-8sintheta)=-15#

Let's solve the second:
#9rcos^2theta+4rsin^2theta=8sintheta-24costheta-15#

#r(9cos^2theta+4sin^2theta)=8sintheta-24costheta-15#

#r=(8sintheta-24costheta-15)/(9cos^2theta+4sin^2theta)#