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

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How do you convert $2 x + y^{2} = 5$ $2x+y^2 = 5$ 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:

Explanation:

Substitute $(r cos (θ))$ $(rcos(theta))$ for x and $r^{2} {sin}^{2} (θ)$ $r^2sin^2(theta)$ for $y^{2}$ $y^2$ :

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

Use the positive root of the quadratic formula:

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

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

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

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

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