How do you convert $- y + 2 = {(3 x + 4)}^{2} + {(y - 1)}^{2}$ $-y+2=(3x+4)^2+(y-1)^2$ into polar form?

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How do you convert $- y + 2 = {(3 x + 4)}^{2} + {(y - 1)}^{2}$ $-y+2=(3x+4)^2+(y-1)^2$ into polar form?

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Answer 1 · 2018-07-01T12:20:10

Given that $- y + 2 = {(3 x + 4)}^{2} + {(y - 1)}^{2}$ $-y+2=(3x+4)^2+(y-1)^2$

Substituting $x = r cos θ$ $x=r\cos\theta$ & $y = r sin θ$ $y=r\sin\theta$

$- r sin θ + 2 = {(3 r cos θ + 4)}^{2} + {(r sin θ - 1)}^{2}$ $-r\sin\theta+2=(3r\cos\theta+4)^2+(r\sin\theta-1)^2$

$- r sin θ + 2 = 9 r^{2} {cos}^{2} θ + 16 + 24 r cos θ + r^{2} {sin}^{2} θ + 1 - 2 r sin θ$ $-r\sin\theta+2=9r^2\cos^2\theta+16+24r\cos\theta+r^2\sin^2\theta+1-2r\sin\theta$

$(8 {cos}^{2} θ + 1) r^{2} + (24 cos θ - sin θ) r + 15 = 0$ $(8\cos^2\theta+1)r^2+(24\cos\theta-\sin\theta)r+15=0$

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How do you convert $- y + 2 = {(3 x + 4)}^{2} + {(y - 1)}^{2}$ $-y+2=(3x+4)^2+(y-1)^2$ into polar form?

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How do you convert $- y + 2 = {(3 x + 4)}^{2} + {(y - 1)}^{2}$ $-y+2=(3x+4)^2+(y-1)^2$ into polar form?