What is 6y+y^2=x^2 in polar form?

Question

#6y+y^2=x^2# in polar form.

I know the answer is #r=(6sintheta)/(cos2theta)#

but I do not know how to get there.

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Answer 1 · 2018-04-24T05:50:04

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

and use the property #cos^2theta-sin^2theta=cos2theta#

#6y+y^2=x^2#........................ (given equation)

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

#6rsintheta +r^2sin^2theta=r^2cos^2theta#

#rArr6rsintheta=r^2cos^2theta-r^2sin^2theta#

#rArr6rsintheta=r^2(cos^2theta-sin^2theta)#

#rArr6rsintheta=r^2cos2theta#..............#{cos^2theta-sin^2theta=cos2theta}#

#rArr6sintheta=rcos2theta#

#:.r=(6sintheta)/(cos2theta)# is the Polar form of #6y+y^2=x^2#