Thanks for helping! This question is really weird :(
r=-6sin theta
I need to multiply both sides of the equation by r and use r^2=x^2+y^2 to rewrite the equation in terms of x and y.

- anonymous

- katieb

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- anonymous

- Astrophysics

Ok so you want to convert it to rectangular, it's good you notice we have to multiply r and use \[r^2 = x^2+y^2\] we know the ratio for sin theta is the following \[\sin \theta = \frac{ y }{ r }\] so we have \[r = - 6\left( \frac{ y }{ r } \right)\] can you finish it off?

- anonymous

I'm not sure I understand where to go from there

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## More answers

- Astrophysics

\[r^2 = - 6y\]

- Astrophysics

What's next?

- anonymous

square - 6y?

- Astrophysics

Why? Look at what we know, and we want to "get rid" of the polar coordinates.

- anonymous

x^2?

- Astrophysics

I don't know what that means

- anonymous

add it in?

- Astrophysics

Hint: \[r^2 = x^2+y^2\]

- anonymous

-6y=x^2+y^2?

- Astrophysics

Yes, that looks good :)

- Astrophysics

You can rearrange it and what not if you wish

- anonymous

That is it in terms of x and y?

- anonymous

I'm asked to complete the square to produce another equation in my worksheet. Should I do it from this form?

- Astrophysics

You may complete the square

- Astrophysics

Ah, yes we have to complete the square, you know how to do that right?

- Astrophysics

When you have it as such \[-6y=x^2+y^2 \] it's always best to complete the square as it will be in terms of x and y.

- anonymous

I know how to do it in regular form, but I'm sort of confused about this one

- Astrophysics

\[-6y = x^2+y^2 \implies x^2+y^2 + 6y = 0\]

- anonymous

Oh, I thought it was when you added (b/2)^2 to both sides or something like that

- Astrophysics

I didn't complete the square...I put in a form so you can complete the square.

- anonymous

Oh, okay gotcha

- Astrophysics

So we don't need to worry about the x^2, now complete the square for y^2+6y

- Astrophysics

When you have \[x^2+ax \implies x^2+ 2 \frac{ ax }{ 2 } + \left( \frac{ a }{ 2 } \right)^2-\left( \frac{ a }{ 2 } \right)^2\] to complete the square.

- anonymous

Wouldn't that make it remain the same?

- anonymous

Sorry, I suck at pre calc :(

- Astrophysics

You should try it, otherwise me doing everything is not going to help you :P

- anonymous

very true

- anonymous

But when plugging in y^2+6y in, it cancels back to y^2+6y, no?

- Astrophysics

Mhm, no you don't do that, we don't plug in anything I just put if you have the form y^2+6y you do the following -> .....etc

- Astrophysics

\[x^2+ax \implies x^2+ 2 \frac{ ax }{ 2 } + \left( \frac{ a }{ 2 } \right)^2-\left( \frac{ a }{ 2 } \right)^2\]
\[x^2+ ax \implies y^2+6y\] in your question

- anonymous

y ^2+6x+9-9?

- anonymous

wait, no

- Astrophysics

|dw:1434416994934:dw| now we just factor

- Astrophysics

|dw:1434417167196:dw|

- Astrophysics

So now we have \[x^2+y^2+6y+9 = -9\] one step away from completing it

- Astrophysics

You were right, but you put 6x instead of 6y :P

- anonymous

Now I factor?

- Astrophysics

That should be = 9 not -9, yes you want it to look as such \[\huge x^2+(y+3)^2 = 9 \] that's your final answer

- anonymous

x^2+(y+3)^2-9=0

- Astrophysics

Good!

- anonymous

oh whoop, same thing ha ha

- anonymous

Thanks for the help!

- anonymous

That really helped

- Astrophysics

No problem, it makes sense now? :)

- anonymous

Yea a lot more!

- Astrophysics

Awesome, glad to hear it, make sure you go over it again!

- anonymous

Of course :), have a good day

- Astrophysics

You to, bye :)

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