Question

transform the following polar equation into an equation in rectangular coordinates r=-4 sin theta

would it be x^2(y+2)^2=4?

Answer 1

\(r = -4\sin\theta\) is a circle

Answer 2

the graph wont change between polar and cartesian

Answer 3

does your equation look anything like the eqn of a circle ?

Answer 4

no so y=-4?

Answer 5

aren't you just guessing the options ?

Answer 6

Remember \[x = r \cos \theta\]\[y= r \sin \theta\]\[r^2 = x^2+y^2 \implies r = \sqrt{x^2+y^2}\]

Answer 7

Multiply both sides by \(r\) to see it.

Answer 8

how would I plug in the numbers though?

Answer 9

\[r^2 = -4rsin \theta \] doing as parth suggested, see what you can do with this given the information above.

Answer 10

\[\color{red}{r^2} = -4\color{blue}{r\sin\theta}\]
you don't want to see \(r, \theta\)

so replace \(\color{red}{r^2} \) by \(\color{red}{x^2+y^2} \)
and \(\color{blue}{r\sin\theta}\) by \(\color{blue}{y}\)


just algebra circus

Answer 11

therefore X+Y=-4? I get it

Answer 12

Not quite..

Answer 13

\[x^2+y^2=-4y\]

\[r^2 = x^2+y^2~~~~y = r \sin \theta\]

Answer 14

And then from there you just substitute right?

Answer 15

technically we're done with the transformation
\(x^2+y^2=-4y\) is the corresponding rectangular equation

are you given options or something ?

Answer 16

I was right actually because x^2(y+2)^2=4 was correct

Answer 17

you're correct upto a + sign

Answer 18

completing the square gives you \(x^2\color{red}{+}(y+2)^2=4\)

which is slightly different from \(x^2(y+2)^2=4\) lexically,
but totally different semantically.

you might think "its just a +"... graph each of them and see

Answer 19

\[x^2+y^2=-4y\]\[x^2+(y^2+4y)=0\]\[x^2+(y^2+4y+\color{red}{4})=4\]\[\boxed{x^2+(y+2)^2 = 4}\]