How do you convert $2 = {(8 x - y)}^{2} - 5 y + 7 x$ $2=(8x-y)^2-5y+7x$ into polar form?

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How do you convert $2 = {(8 x - y)}^{2} - 5 y + 7 x$ $2=(8x-y)^2-5y+7x$ into polar form?

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Answer 1 · 2016-01-18T21:56:20

$(64 x^{2} - 16 x y + y^{2}) - 5 y + 7 x - 2 = 0$ $(64x^2- 16xy+y^2)-5y+7x-2=0$
Now sub:
$x = r cos θ; y = r sin θ$ $x= rcostheta; y=rsin theta$
Noting that $r = x^{2} + y^{2}; θ = {tan}^{- 1} (\frac{y}{x})$ $r=x^2+y^2; theta=tan^-1 (y/x)$

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How do you convert $2 = {(8 x - y)}^{2} - 5 y + 7 x$ $2=(8x-y)^2-5y+7x$ into polar form?

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How do you convert 2=(8x-y)^2-5y+7x2=(8x−y)2−5y+7x2=(8x-y)^2-5y+7x into polar form?

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How do you convert $2 = {(8 x - y)}^{2} - 5 y + 7 x$ $2=(8x-y)^2-5y+7x$ into polar form?