How do you convert $x^{2} = 8 y$ $x^2 = 8y$ into polar form?

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Answer 1 · 2015-11-21T11:48:27

$r = 0$ $r = 0$ OR $r = \frac{8 sin θ}{{cos}^{2} θ}$ $r = (8 sin theta) / cos^2 theta$

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

$r^{2} {cos}^{2} θ = 8 r sin θ$ $r^2 cos^2 theta = 8 r sin theta$

$r = 0$ $r = 0$ OR $r = \frac{8 sin θ}{{cos}^{2} θ}$ $r = (8 sin theta) / cos^2 theta$

what if $cos θ = 0$ $cos theta = 0$ ?

$r^{2} 0 = 8 r (\pm 1) \Rightarrow r = 0$ $r^2 0 = 8r (±1) Rightarrow r = 0$

How do you convert x^2 = 8yx2=8yx^2 = 8y into polar form?