How do you convert $y^{2} = 9 x$ $y^2 =9x$ to polar form?

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How do you convert $y^{2} = 9 x$ $y^2 =9x$ to polar form?

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Answer 1 · 2018-04-01T00:11:46

$r = 9 cot θ csc θ$ $r=9cotthetacsctheta$

Explanation:

The conversion from Rectangular to Polar:
$x = r cos θ$ $x=rcostheta$
$y = r sin θ$ $y=rsintheta$

Substitute for $x$ $x$ and $y$ $y$ :
${(r sin θ)}^{2} = 9 (r cos θ)$ $(rsintheta)^2=9(rcostheta)$
$r^{2} {sin}^{2} θ = 9 r cos θ$ $r^2sin^2theta=9rcostheta$
$r^{2} {sin}^{2} θ - 9 r cos θ = 0$ $r^2sin^2theta-9rcostheta=0$
$r (r {sin}^{2} θ - 9 cos θ) = 0$ $r(rsin^2theta-9costheta)=0$

At this point either $r = 0$ $r=0$ or $r {sin}^{2} θ - 9 cos θ = 0$ $rsin^2theta-9costheta=0$ , let's solve the second one to get a meaningful answer:

$r {sin}^{2} θ = 9 cos θ$ $rsin^2theta=9costheta$

$r = \frac{9 cos θ}{{sin}^{2} θ}$ $r=(9costheta)/sin^2theta$

$r = \frac{9 cos θ}{sin θ} \cdot \frac{1}{sin θ}$ $r= (9costheta)/sintheta*1/sintheta$

Remember: $\frac{cos θ}{sin θ} = cot θ$ $costheta/sintheta=cottheta$ and $\frac{1}{sin θ} = csc θ$ $1/sintheta= csctheta$ :

$r = 9 cot θ csc θ$ $r=9cotthetacsctheta$

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How do you convert $y^{2} = 9 x$ $y^2 =9x$ to polar form?

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