conver the polar equation to rectangular form.

Question

anonymous · Answer

2/2-sin theta

anonymous · Answer

r=2/2-sin theta

Here is what I am thinking:
r = 2/(2-y/r)
r(2-y/r) = 2
2r - y =2
2r = 2+y
r+ 1 + y/2

anonymous · Answer

it should be r = 1+ y/2  ???

terenzreignz · Answer

As for you, I thought you were doing great, up until the step after
2r = 2+y
Up to here, was good... how'd you get the next step?

anonymous · Answer

I divided everything by 2

terenzreignz · Answer

ahh.. a typo on your part, aye? Okay... the first + sign should have been an = sign.

anonymous · Answer

yup ;)

terenzreignz · Answer

Now use this relation... we don't want any r's if it's going to be a rectangular form, so

$$\Large r = \sqrt{x^2+y^2}$$

anonymous · Answer

from the 2r-y=2?

terenzreignz · Answer

Yeah, sure :)
Better from
2r = y + 2 
though

anonymous · Answer

ok. should I be plugging that info into that equation?

terenzreignz · Answer

It then becomes
$$\large 2\sqrt{x^2+y^2}= y + 2$$

terenzreignz · Answer

It's not pretty, so let's square both sides...
$$\Large 4(x^2+y^2) = (y+2)^2$$

anonymous · Answer

Oh!  Okay that makes a bit more sense.  Would this involve another step like factoring it out?