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

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Answer 1 · 2018-03-16T09:18:44

$4 r^{2} (1 + sin 2 θ) + r (7 sin θ - 4 cos θ) - 9 = 0$ $4r^2(1+sin2theta)+r(7sintheta-4costheta)-9=0$

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

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

$9 = {(2 x + 2 y)}^{2} + 7 y - 4 x$ $9=(2x+2y)^2+7y-4x$ can be written as

$9 = 4 x^{2} + 4 y^{2} + 8 x y + 7 y - 4 x$ $9=4x^2+4y^2+8xy+7y-4x$

or $9 = 4 r^{2} + 8 r^{2} sin θ cos θ + 7 r sin θ - 4 r cos θ$ $9=4r^2+8r^2sinthetacostheta+7rsintheta-4rcostheta$

or $4 r^{2} (1 + 2 sin θ cos θ) + r (7 sin θ - 4 cos θ) - 9 = 0$ $4r^2(1+2sinthetacostheta)+r(7sintheta-4costheta)-9=0$

or $4 r^{2} (1 + sin 2 θ) + r (7 sin θ - 4 cos θ) - 9 = 0$ $4r^2(1+sin2theta)+r(7sintheta-4costheta)-9=0$

graph{9=(2x+2y)^2+7y-4x [-4.083, 5.917, -3.6, 1.4]}

How do you convert 9=(2x+2y)^2+7y-4x9=(2x+2y)2+7y−4x9=(2x+2y)^2+7y-4x into polar form?