How do you convert $6 = {(5 x - 7 y)}^{2} - y + x$ $6=(5x-7y)^2-y+x$ into polar form?

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How do you convert $6 = {(5 x - 7 y)}^{2} - y + x$ $6=(5x-7y)^2-y+x$ into polar form?

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Answer 1 · 2018-04-15T03:58:28

$2 r^{2} sin θ (12 sin θ - 35 cos θ) + r (cos θ - sin θ) + 19 = 0$ $color(indigo)(2r^2 sin theta (12 sin theta - 35 cos theta ) + r (cos theta - sin theta) + 19 = 0$

Explanation:

$x = r cos θ, y = r sin θ$ $x = r cos theta, y = r sin theta$

${(5 x - 7 y)}^{2} - y + x = 6$ $(5x - 7y)^2 - y + x = 6$

$Substituting values of x & y in terms of r and θ$ $"Substituting values of x & y in terms of " r and theta$ ,

${(5 r cos θ - 7 r sin θ)}^{2} - r sin θ + cos θ = 6$ $(5 r cos theta - 7 r sin theta)^2 - r sin theta + cos theta = 6$

$25 r^{2} {cos}^{2} θ + 49 r^{2} {sin}^{2} θ - 70 r^{2} cos θ sin θ - r sin θ + r cos θ = 6$ $25r^2 cos^2 theta + 49 r^2 sin^2 theta - 70 r^2 cos theta sin theta - r sin theta + r cos theta = 6$

$25 r^{cos}^2 θ + 25 r^{2} {sin}^{2} θ + 24 r^{sin}^2 θ - 70 r^{2} cos θ sin θ - r sin θ + r cos θ = 6$ $25 r^ cos^2 theta + 25 r^2 sin^2 theta + 24 r^ sin^2 theta - 70 r^2 cos theta sin theta - r sin theta + r cos theta = 6$

$25 + 24 r^{2} {sin}^{2} θ - 70 r^{2} cos θ sin θ - r sin θ + r cos θ = 6$ $25 + 24 r^2 sin^2 theta - 70 r^2 cos theta sin theta - r sin theta + r cos theta = 6$

$2 r^{2} sin θ (12 sin θ - 35 cos θ) + r (cos θ - sin θ) + 19 = 0$ $2r^2 sin theta (12 sin theta - 35 cos theta ) + r (cos theta - sin theta) + 19 = 0$

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How do you convert $6 = {(5 x - 7 y)}^{2} - y + x$ $6=(5x-7y)^2-y+x$ into polar form?

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How do you convert 6=(5x-7y)^2-y+x6=(5x−7y)2−y+x6=(5x-7y)^2-y+x into polar form?

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How do you convert $6 = {(5 x - 7 y)}^{2} - y + x$ $6=(5x-7y)^2-y+x$ into polar form?