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

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Answer 1 · 2018-08-06T08:40:11

$8 r^{2} - 2 r sin θ - (5 cos θ - sin θ) = 0$ $8 r^2 - 2 r sin theta - ( 5 cos theta - sin theta ) = 0$ .

Conversion formulas:

$(x, y) = r (cos θ, sin θ), r = \sqrt{x^{2} + y^{2}} \geq 0$ $( x, y ) = r ( cos theta,sin theta ), r = sqrt ( x^2 + y^2 ) >= 0$

$8 = {(5 x - y)}^{2} + 2 y$ $8 = ( 5 x - y )^2 + 2y$ converts to

$8 = \frac{1}{r^{2}} (5 cos θ - sin θ) + \frac{2}{r} sin θ$ $8 = 1/r^2 ( 5 cos theta - sin theta ) + 2/r sin theta$ , giving

$8 r^{2} - 2 r sin θ - (5 cos θ - sin θ) = 0$ $8 r^2 - 2 r sin theta - ( 5 cos theta - sin theta ) = 0$ .

See this parabola.
graph{(5x-y)^2+2y-8=0}

How do you convert 8=(5x-y)^2+2y8=(5x−y)2+2y8=(5x-y)^2+2y into polar form?