## anonymous 5 years ago Convert to polar form: $4x^{2}-5y^{2}-36y-36=0$

1. anonymous

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

2. anonymous

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

4. watchmath

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

show = saw

6. anonymous

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

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

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

9. watchmath

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

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