How do you convert $9 = {(4 x - 2 y)}^{2} - y - 6$ $9=(4x-2y)^2-y-6$ into polar form?

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Answer 1 · 2018-05-21T00:11:10

$r^{2} {(4 cos θ - 2 sin θ)}^{2} - r sin θ = 15$ $r^2 (4 cos theta - 2 sin theta)^2 - r sin theta = 15$

$9 = {(4 x - 2 y)}^{2} - y - 6$ $9=(4x-2y)^2-y-6$

We just substitute $x = r cos θ$ $x=r cos theta$ and $y = r sin θ$ $y=r sin theta$

$9 = {(4 r cos θ - 2 r sin θ)}^{2} - r sin θ - 6$ $9 = (4r cos theta - 2 r sin theta)^2 - r sin theta -6$

$r^{2} {(4 cos θ - 2 sin θ)}^{2} - r sin θ = 15$ $r^2 (4 cos theta - 2 sin theta)^2 - r sin theta = 15$

The original equation we're given wasn't simplified so it's probably OK to stop here.

How do you convert 9=(4x-2y)^2-y-69=(4x−2y)2−y−69=(4x-2y)^2-y-6 into polar form?