How do you convert $9 = {(4 x + 7)}^{2} + {(6 y - 9)}^{2}$ $9=(4x+7)^2+(6y-9)^2$ into polar form?

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Answer 1 · 2016-05-19T17:31:41

$r^{2} (4 + 5 {sin}^{2} θ) + r (14 cos θ - 27 sin θ) + 18 = 0$ $r^2(4+5sin^2theta)+r(14costheta-27sintheta)+18=0$

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

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

Using them ${(4 x + 7)}^{2} + {(6 y - 9)}^{2} = 9$ $(4x+7)^2+(6y-9)^2=9$ is

${(4 x + 7)}^{2} + {(6 y - 9)}^{2} = 9$ $(4x+7)^2+(6y-9)^2=9$

${(4 r cos θ + 7)}^{2} + {(6 r sin θ - 9)}^{2} = 9$ $(4rcostheta+7)^2+(6rsintheta-9)^2=9$

or $16 r^{2} {cos}^{2} θ + 56 r cos θ + 36 r^{2} {sin}^{2} θ - 108 r sin θ + 81 = 9$ $16r^2cos^2theta+56rcostheta+36r^2sin^2theta-108rsintheta+81=9$

or $16 r^{2} + 56 r cos θ + 20 r^{2} {sin}^{2} θ - 108 r sin θ + 72 = 0$ $16r^2+56rcostheta+20r^2sin^2theta-108rsintheta+72=0$

or $4 r^{2} + 14 r cos θ + 5 r^{2} {sin}^{2} θ - 27 r sin θ + 18 = 0$ $4r^2+14rcostheta+5r^2sin^2theta-27rsintheta+18=0$

or $r^{2} (4 + 5 {sin}^{2} θ) + r (14 cos θ - 27 sin θ) + 18 = 0$ $r^2(4+5sin^2theta)+r(14costheta-27sintheta)+18=0$

How do you convert 9=(4x+7)^2+(6y-9)^29=(4x+7)2+(6y−9)29=(4x+7)^2+(6y-9)^2 into polar form?