How do you convert #9=(2x+2y)^2+7y-4x# into polar form?

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Answer 1 · 2018-03-16T09:18:44

#4r^2(1+sin2theta)+r(7sintheta-4costheta)-9=0#

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

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

#9=(2x+2y)^2+7y-4x# can be written as

#9=4x^2+4y^2+8xy+7y-4x#

or #9=4r^2+8r^2sinthetacostheta+7rsintheta-4rcostheta#

or #4r^2(1+2sinthetacostheta)+r(7sintheta-4costheta)-9=0#

or #4r^2(1+sin2theta)+r(7sintheta-4costheta)-9=0#

graph{9=(2x+2y)^2+7y-4x [-4.083, 5.917, -3.6, 1.4]}