So I am not sure how to do this problem : "Convert rectangular equation $(x^2+y^2)^2 - 4(x^2-y^2) = 0$ to polar form". Any help would be appreciated.
Convert rectangular equation $(x^2+y^2)^2 - 4(x^2-y^2) = 0$ to polar form
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$\begingroup$
polar-coordinates
parametric
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0**HINT:** Recall that $x = r\cos(\theta)$ and $y = r\sin(\theta)$ (moreover, $r^2 = x^2 + y^2$, which can be handy) – 2017-02-07
1 Answers
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$$0=(x^2+y^2)^2 - 4(x^2-y^2) $$ Recall that $x^2+y^2=r^2$ and $x=r\cos(\theta)$ and $y=r\sin(\theta)$
$$ 0=(r^2)^2-4(r^2 \cos^2(\theta)-r^2\sin^2(\theta))$$
$$ 0=r^4-4(r^2(\cos^2(\theta)-\sin^2(\theta))$$
$$ 0 = r^4-4r^2\cos(2\theta)$$
$$ 0 = r^2(r^2-4\cos(2\theta))$$
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1you can go further! – 2017-02-07
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0@Arnaldo Oh I thought the zero was an "O"... – 2017-02-07
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0@Arnaldo is it r = sqrt(4cos(2theta)) – 2017-02-07
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0@Nahian: you can leave like the answer or you can split in $r=0$ or what you suggested. – 2017-02-07