anonymous
  • anonymous
transform the following polar equation into an equation in rectangular coordinates r=-4 sin theta would it be x^2(y+2)^2=4?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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ganeshie8
  • ganeshie8
\(r = -4\sin\theta\) is a circle
ganeshie8
  • ganeshie8
the graph wont change between polar and cartesian
ganeshie8
  • ganeshie8
does your equation look anything like the eqn of a circle ?

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More answers

anonymous
  • anonymous
no so y=-4?
ganeshie8
  • ganeshie8
aren't you just guessing the options ?
anonymous
  • anonymous
Remember \[x = r \cos \theta\]\[y= r \sin \theta\]\[r^2 = x^2+y^2 \implies r = \sqrt{x^2+y^2}\]
ParthKohli
  • ParthKohli
Multiply both sides by \(r\) to see it.
anonymous
  • anonymous
how would I plug in the numbers though?
anonymous
  • anonymous
\[r^2 = -4rsin \theta \] doing as parth suggested, see what you can do with this given the information above.
ganeshie8
  • ganeshie8
\[\color{red}{r^2} = -4\color{blue}{r\sin\theta}\] 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
anonymous
  • anonymous
therefore X+Y=-4? I get it
anonymous
  • anonymous
Not quite..
anonymous
  • anonymous
\[x^2+y^2=-4y\] \[r^2 = x^2+y^2~~~~y = r \sin \theta\]
anonymous
  • anonymous
And then from there you just substitute right?
ganeshie8
  • ganeshie8
technically we're done with the transformation \(x^2+y^2=-4y\) is the corresponding rectangular equation are you given options or something ?
anonymous
  • anonymous
I was right actually because x^2(y+2)^2=4 was correct
ganeshie8
  • ganeshie8
you're correct upto a + sign
ganeshie8
  • ganeshie8
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
Jhannybean
  • Jhannybean
\[x^2+y^2=-4y\]\[x^2+(y^2+4y)=0\]\[x^2+(y^2+4y+\color{red}{4})=4\]\[\boxed{x^2+(y+2)^2 = 4}\]
triciaal
  • triciaal
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UsukiDoll
  • UsukiDoll
the answer is right. The user just forgot to add the + sign during the final stages of this problem. \[x^2\color{red}{+}(y+2)^2=4 \]

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