How do you convert #9=(-2x+y)^2-5y+3x# into polar form?

1 Answer
1s2s2p
Mar 14, 2018

#9=4r^2cos^2(theta)-4r^2sinthetacostheta+r^2sin^2(theta)-5rsintheta+3rcostheta=r(sintheta(r(sintheta−4costheta)−5)+costheta(4rcostheta+3))#

Explanation:

#x=rcostheta#
#y=rsintheta#

#9=(-2(rcostheta)+rsintheta)^2-5rsintheta+3rcostheta#

#9=4r^2cos^2(theta)-4r^2sinthetacostheta+r^2sin^2(theta)-5rsintheta+3rcostheta#

#9=r(sintheta(r(sintheta−4costheta)−5)+costheta(4rcostheta+3))#

Related questions
See all questions in Converting Between Systems
Impact of this question
1294 views around the world
You can reuse this answer
Creative Commons License