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

6y+y^2=x^26y+y2=x2 in polar form.

I know the answer is r=(6sintheta)/(cos2theta)r=6sinθcos2θ

but I do not know how to get there.

1 Answer
Kalit Gautam
Apr 24, 2018

Put x=rcostheta x=rcosθ and y=rsinthetay=rsinθ

and use the property cos^2theta-sin^2theta=cos2thetacos2θ−sin2θ=cos2θ

Explanation:

6y+y^2=x^26y+y2=x2........................ (given equation)

Put x=rcostheta x=rcosθ and y=rsinthetay=rsinθ , we get :-

6rsintheta +r^2sin^2theta=r^2cos^2theta6rsinθ+r2sin2θ=r2cos2θ

rArr6rsintheta=r^2cos^2theta-r^2sin^2theta⇒6rsinθ=r2cos2θ−r2sin2θ

rArr6rsintheta=r^2(cos^2theta-sin^2theta)⇒6rsinθ=r2(cos2θ−sin2θ)

rArr6rsintheta=r^2cos2theta⇒6rsinθ=r2cos2θ..............{cos^2theta-sin^2theta=cos2theta}{cos2θ−sin2θ=cos2θ}

rArr6sintheta=rcos2theta⇒6sinθ=rcos2θ

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

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