Question

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

Answer 1

x=rcos(theta)
y=rsin(theta)

substitute.

Answer 2

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?

Answer 3

Help please!

Answer 4

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.

Answer 5

show = saw

Answer 6

Thank you. But the problem is the answer is multiple choice, and the options are:

\[A)\quad r=\frac{-4}{1+\sin\theta}\]\[B)\quad r=\frac{-4}{1+\cos\theta}\]\[C)\quad r=\frac{6}{1-\sin\theta}\]\[D)\quad r=\frac{6}{2-3\sin\theta}\]\[E)\quad None\;of\;the\;above.\]

Answer 7

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?

Answer 8

i did it watchmath :) come grade me :) when you get a chance

Answer 9

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

Answer 10

Thank you. I really appreciate your generosity in answering so many questions. Do you mind if I ask you what motivates you?