How do you convert $1 = {(x + 4)}^{2} + {(7 y - 2)}^{2}$ $1=(x+4)^2+(7y-2)^2$ into polar form?

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How do you convert $1 = {(x + 4)}^{2} + {(7 y - 2)}^{2}$ $1=(x+4)^2+(7y-2)^2$ into polar form?

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Answer 1 · 2016-10-04T09:24:18

$r^{2} (1 + 48 {sin}^{2} θ) + r (8 cos θ - 28 sin θ) + 19 = 0$ $r^2(1+48sin^2theta)+r(8costheta-28sintheta)+19=0$

Explanation:

The relation between rectangular Cartesian coordinates $(x, y)$ $(x,y)$ and polar coordinates $(r, θ)$ $(r,theta)$ is given by

$x = r cos θ$ $x=rcostheta$ , $y = r sin θ$ $y=rsintheta$ and $x^{2} + y^{2} = r^{2}$ $x^2+y^2=r^2$

Hence $y^{2} + {(x - 3)}^{2} = 9$ $y^2+(x-3)^2=9$ can be rewritten as

${(x + 4)}^{2} + {(7 y - 2)}^{2} = 1$ $(x+4)^2+(7y-2)^2=1$

or ${(r cos θ + 4)}^{2} + {(7 r sin θ - 2)}^{2} = 1$ $(rcostheta+4)^2+(7rsintheta-2)^2=1$

or $(r^{2} {cos}^{2} θ + 8 r cos θ + 16) + (49 r^{2} {sin}^{2} θ - 28 r sin θ + 4) = 1$ $(r^2cos^2theta+8rcostheta+16)+(49r^2sin^2theta-28rsintheta+4)=1$

or $r^{2} {cos}^{2} θ + 8 r cos θ + 49 r^{2} {sin}^{2} θ - 28 r sin θ + 20 - 1 = 0$ $r^2cos^2theta+8rcostheta+49r^2sin^2theta-28rsintheta+20-1=0$

or $r^{2} (1 + 48 {sin}^{2} θ) + r (8 cos θ - 28 sin θ) + 19 = 0$ $r^2(1+48sin^2theta)+r(8costheta-28sintheta)+19=0$

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How do you convert $1 = {(x + 4)}^{2} + {(7 y - 2)}^{2}$ $1=(x+4)^2+(7y-2)^2$ into polar form?

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Explanation:

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