How do you convert $2x+y^2 = 5$ to polar form?

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Answer 1 · 2016-11-13T03:30:02

Please see the explanation for steps leading to the polar form:

Substitute $(rcos(theta))$ for x and $r^2sin^2(theta)$ for $y^2$ :

$r^2sin^2(theta) +$ 2rcos(theta) - 5 = 0#

Use the positive root of the quadratic formula:

$r = {-2cos(theta) + sqrt(4cos^2(theta) + 20sin^2(theta))}/(2sin^2(theta))$

$r = {-cos(theta) + sqrt(cos^2(theta) + 5sin^2(theta))}/(sin^2(theta))$

$r = {-cos(theta) + sqrt(1 + 4sin^2(theta))}/(sin^2(theta))$

How do you convert 2x+y^2 = 52x+y^2 = 5 to polar form?