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

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

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Answer 1 · 2016-06-05T13:24:33

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

Explanation:

Use $(x, y) = (r cos θ, r sin θ)$ $(x, y) = (r cos theta, r sin theta)$

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

$r = 0 and r = 8 cot θ csc θ$ $r =0 and r = 8 cot theta csc theta$

The second equation inludes r = 0, at $θ = \pm \frac{π}{2}$ $theta =+-pi/2$ . .

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

from $\frac{π}{2}$ $pi/2$ to 0, backwards,. $and θ \in (- \frac{π}{2}, 0)$ $and theta in (-pi/2, 0)$ .

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

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