How do you convert $Y = 4 x^{2}$ $Y=4x^2$ into polar form?

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Answer 1 · 2017-09-12T18:10:06

see below

Use the formula $x = r cos θ, y = r sin θ$ $x = r cos theta, y = r sin theta$

Hence,

$y = 4 x^{2}$ $y=4x^2$

$r sin θ = 4 r^{2} {cos}^{2} θ$ $r sin theta = 4r^2 cos^2 theta$

$:. 0=4r^2 cos^2 theta -r sin theta$

How do you convert Y=4x^2Y=4x2Y=4x^2 into polar form?