How do you convert $2 = {(8 x - 7 y)}^{2} - 7 x$ $2=(8x-7y)^2-7x$ into polar form?

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How do you convert $2 = {(8 x - 7 y)}^{2} - 7 x$ $2=(8x-7y)^2-7x$ into polar form?

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Answer 1 · 2018-06-11T09:06:40

$\Rightarrow (15 {sin}^{2} θ - 56 sin 2 θ + 49) r^{2} - 2 = 0$ $=>(15sin^2theta-56sin2theta+49)r^2-2=0$

Explanation:

put $x = r cos θ, y = r sin θ$ $x=rcostheta,y=rsintheta$ and solve
$2 = {(8 x - 7 y)}^{2} - 7 x$ $2=(8x-7y)^2-7x$
$\Rightarrow 2 = 64 x^{2} + 49 y^{2} - 112 x y - 7 x$ $=>2=64x^2+49y^2-112xy-7x$
$\Rightarrow 2 = 49 (x^{2} + y^{2}) + 15 x^{2} - 7 x - x y$ $=>2=49(x^2+y^2)+15x^2-7x-xy$
$\Rightarrow 2 = 49 r^{2} ({sin}^{2} θ + {cos}^{2} θ) + 15 r^{2} {sin}^{2} θ - 56.2 r^{2} sin θ cos θ$ $=>2=49r^2(sin^2theta+cos^2theta)+15r^2sin^2theta-56.2r^2sinthetacostheta$
$\Rightarrow 2 = 49 r^{2} + 15 r^{2} {sin}^{2} θ - 56 r^{2} sin 2 θ$ $=>2=49r^2+15r^2sin^2theta-56r^2sin2theta$
$\Rightarrow (15 {sin}^{2} θ - 56 sin 2 θ + 49) r^{2} - 2 = 0$ $=>(15sin^2theta-56sin2theta+49)r^2-2=0$

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How do you convert $2 = {(8 x - 7 y)}^{2} - 7 x$ $2=(8x-7y)^2-7x$ into polar form?

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Explanation:

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How do you convert 2=(8x-7y)^2-7x2=(8x−7y)2−7x2=(8x-7y)^2-7x into polar form?

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Explanation:

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How do you convert $2 = {(8 x - 7 y)}^{2} - 7 x$ $2=(8x-7y)^2-7x$ into polar form?