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Mathematics 16 Online
OpenStudy (anonymous):

Need help with "Polar Equations of Conics" Medal & Fan!

OpenStudy (anonymous):

OpenStudy (anonymous):

c

OpenStudy (loser66):

To me, it is b

OpenStudy (anonymous):

porque

OpenStudy (loser66):

\(r=\dfrac{-72}{12+6sin \theta}=\dfrac{-12}{2+sin\theta}\)\(\implies r(2+sin\theta)=-12\) \(2r +rsin\theta =-12\\2r =-(12+y)\) square both sides \(4r^2 = 144+24y +y^2\\4x^2+4y^2=144+24y+y^2\) Which is \(4x^2+3y^2 -24y \huge\color{red}{-}144\)

OpenStudy (loser66):

=0, it is b

OpenStudy (anonymous):

well i guess you proved me wrong .. good job

OpenStudy (loser66):

I am trying to solve the problem and get that answer. I don't care whether you are wrong or not. :)

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