How do you convert $x^{2} - 12 y - 36 = 0$ $x^2 -12y-36=0$ to polar form?

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How do you convert $x^{2} - 12 y - 36 = 0$ $x^2 -12y-36=0$ to polar form?

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Answer 1 · 2017-11-20T13:00:13

Using $x = r cos θ$ $x=rcostheta$ and $y = r sin θ$ $y=rsintheta$

Explanation:

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Remembering that an x,y coordinate can be represented by a right triangle with angle theta and hypotenuse r, we can derive that $x = r cos θ$ $x=rcostheta$ and $y = r sin θ$ $y=rsintheta$ .

Substituting these into your equation leaves the result
$r^{2} {cos}^{2} θ - 12 r sin θ - 36 = 0$ $r^2cos^2theta-12rsintheta-36=0$

This is a quadratic expression in r, so use the quadratic formula to solve for r
$r = \frac{12 sin θ \pm \sqrt{12^{2} {sin}^{2} θ + 4 \cdot 36 \cdot {cos}^{2} θ}}{2 {cos}^{2} θ}$ $r=(12sintheta+-sqrt(12^2sin^2theta+4*36*cos^2theta))/(2cos^2theta)$
$r = \frac{12 sin θ \pm \sqrt{12^{2} {sin}^{2} θ \pm 12^{2} {cos}^{2} θ}}{2 {cos}^{2} θ}$ $r=(12sintheta+-sqrt(12^2sin^2theta+-12^2cos^2theta))/(2cos^2theta)$
$r = \frac{12 sin θ \pm 12}{2 {cos}^{2} θ}$ $r=(12sintheta+-12)/(2cos^2theta)$

Therefore,
$r = \frac{6 sin θ \pm 6}{{cos}^{2} θ}$ $r=(6sintheta+-6)/cos^2theta$

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How do you convert $x^{2} - 12 y - 36 = 0$ $x^2 -12y-36=0$ to polar form?

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How do you convert $x^{2} - 12 y - 36 = 0$ $x^2 -12y-36=0$ to polar form?