How do you convert #3=(7x-5y)^2-y# into polar form?

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Answer 1 · 2018-04-23T09:37:58

Put #x=rcostheta # and #y=rsintheta#

#3=(7x-5y)^2 - y# ...............(Given equation)

Put #x=rcostheta # and #y=rsintheta# ; we get :-

#rArr3=(7rcostheta-5rsintheta)^2-rsintheta#

#rArr3=49r^2cos^2theta+25r^2sin^2theta-70r^2costheta.sintheta-rsintheta#

#rArr3=24r^2cos^2theta+(25r^2cos^2theta+25r^2sin^2theta)-70r^2costheta.sintheta-rsintheta#

#rArr3=24r^2cos^2theta+25r^2-70r^2costheta.sintheta-rsintheta#

#:.25r^2+24r^2cos^2theta-70r^2costheta.sintheta-rsintheta-3=0#

is the Polar form