How do you convert Y^2=8x into polar form?

Jun 5, 2016

$r = 8 \cot \theta \csc \theta$. To trace the parabola, it is sufficient to vary $\theta$ from $\frac{\pi}{2}$ to 0, backwards.$\mathmr{and} \theta \in \left(- \frac{\pi}{2} , 0\right)$.

Explanation:

Use $\left(x , y\right) = \left(r \cos \theta , r \sin \theta\right)$

Here, ${r}^{2} {\sin}^{2} \theta = 8 r \cos \theta$.

$r = 0 \mathmr{and} r = 8 \cot \theta \csc \theta$

The second equation inludes r = 0, at $\theta = \pm \frac{\pi}{2}$. .

To trace the parabola in the polar mode, it is sufficient to vary $\theta$

from $\frac{\pi}{2}$ to 0, backwards,.$\mathmr{and} \theta \in \left(- \frac{\pi}{2} , 0\right)$.