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 $6y+y^2=x^2$