How do you convert $9 = {(5 x + y)}^{2} - 2 y + x$ $9=(5x+y)^2-2y+x$ into polar form?

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Answer 1 · 2018-05-14T08:20:27

$r = \frac{9}{r {(5 cos θ + sin θ)}^{2} - 2 sin θ + cos θ}$ $r=9/(r(5costheta+sintheta)^2-2sintheta+costheta)$

For this we will need:
$x = r cos θ$ $x=rcostheta$
$y = r sin θ$ $y=rsintheta$

Substituting these equations gives us:
$9 = {(5 r cos θ + r sin θ)}^{2} - 2 r sin θ + r cos θ$ $9=(5rcostheta+rsintheta)^2-2rsintheta+rcostheta$

$9 = r^{2} {(5 cos θ + sin θ)}^{2} - 2 r sin θ + r cos θ$ $9=r^2(5costheta+sintheta)^2-2rsintheta+rcostheta$

$9 = r (r {(5 cos θ + sin θ)}^{2} - 2 sin θ + cos θ)$ $9=r(r(5costheta+sintheta)^2-2sintheta+costheta)$

$r = \frac{9}{r {(5 cos θ + sin θ)}^{2} - 2 sin θ + cos θ}$ $r=9/(r(5costheta+sintheta)^2-2sintheta+costheta)$

How do you convert 9=(5x+y)^2-2y+x9=(5x+y)2−2y+x9=(5x+y)^2-2y+x into polar form?