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anonymous

  • 5 years ago

Convert to polar form: \[4x^{2}-5y^{2}-36y-36=0\]

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  1. anonymous
    • 5 years ago
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    x=rcos(theta) y=rsin(theta) substitute.

  2. anonymous
    • 5 years ago
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    So I began with:\[4(r\cos\theta)^{2}-5(r\sin\theta)^{2}-36(r\sin\theta)=36\]and end up with:\[r\left[(4-9\sin^{2}\theta)r-36\sin\theta\right]=36\] Is this correct? If yes, what is next?

  3. anonymous
    • 5 years ago
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    Help please!

  4. watchmath
    • 5 years ago
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    I show your update on my website. Let me just answer you here. \(4r^2\cos^2\theta-5r^2\sin^2\theta-36r\sin\theta-36=0\) Well actually if you just answer this then nobody can say that this is wrong.

  5. watchmath
    • 5 years ago
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    show = saw

  6. anonymous
    • 5 years ago
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    Thank you. But the problem is the answer is multiple choice, and the options are: \[A)\quad r=\frac{-4}{1+\sin\theta}\]\[B)\quad r=\frac{-4}{1+\cos\theta}\]\[C)\quad r=\frac{6}{1-\sin\theta}\]\[D)\quad r=\frac{6}{2-3\sin\theta}\]\[E)\quad None\;of\;the\;above.\]

  7. anonymous
    • 5 years ago
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    So I put all of these equations on wolframalpha.com and the one that is correct is D. Does anybody know how to get there?

  8. myininaya
    • 5 years ago
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    i did it watchmath :) come grade me :) when you get a chance

  9. watchmath
    • 5 years ago
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    The answer is D From above we have \(4r^2\cos^2\theta-5r^2\sin^2\theta-36r\sin\theta-36=0\) \(4r^2\cos^2\theta=5r^2\sin^2\theta+36r\sin\theta+36\) \(4r^2\cos^2\theta+4r^2\sin^2\theta=9r^2\sin^2\theta+36r\sin\theta+36\) \(4r^2(\cos^2\theta+\sin^2\theta)=(3r\sin\theta+6)^2\) \(4r^2=(3r\sin\theta+6)^2\) \(2r=3r\sin\theta+6\) \((2-3\sin\theta)r=6\) \(r=\frac{6}{2-3\sin\theta}\).

  10. anonymous
    • 5 years ago
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    Thank you. I really appreciate your generosity in answering so many questions. Do you mind if I ask you what motivates you?

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