Question

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

`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.

Answer 1

Put `x=rcostheta` and `y=rsintheta`

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

Explanation:

`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`