How do you convert #8=(6x-4y)^2+2y-x# into polar form?

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Answer 1 · 2018-01-01T15:54:35

#r^2(10cos2theta-24sin2theta+6)+r(2sintheta-costheta)-8=0#

The relation between polar coordinates #(r,theta)# and Cartesian coordinates #(x,y)# is #x=rcostheta#, #y=rsintheta# and #x^2+y^2=r^2#.

Hence #8=(6x-4y)^2+2y-x# can be written as

#8=36x^2+16y^2-48xy+2y-x#

or #8=36r^2cos^2theta+16r^2sin^2theta-48r^2sinthetacostheta+2rsintheta-rcostheta#

or #8=20r^2cos^2theta+16r^2-24r^2sin2theta+2rsintheta-rcostheta#

or #r^2(20cos^2theta-24sin2theta+16)+r(2sintheta-costheta)-8=0#

or #r^2(10(cos2theta-1)-24sin2theta+16)+r(2sintheta-costheta)-8=0#

or #r^2(10cos2theta-24sin2theta+6)+r(2sintheta-costheta)-8=0#