How do you convert y=3x-x^2y=3x−x2 into polar form?

2 Answers
A. S. Adikesavan
Feb 12, 2016

x= r cos(theta), y = r sin(theta)x=rcos(θ),y=rsin(θ). Answer is r = sec(theta) ( 3 - tan(theta))r=sec(θ)(3−tan(θ)).

Explanation:

The equation in rectangular form represents a parabola with vertex at (3/2, 9/4) and axis along negative y\axis. The parabola does not pass through the origin. So, r is never 0.

Jim G.
Feb 12, 2016

r = sectheta(3- tantheta ) r=secθ(3−tanθ)

Explanation:

using the formulae that links Cartesian to Polar coordinates.

• x = rcostheta ∙x=rcosθ

• y = rsintheta ∙y=rsinθ

the question can then be written as:

rsintheta = 3rcostheta - r^2cos^2theta rsinθ=3rcosθ−r2cos2θ

hence r^2cos^2theta = 3rcostheta - rsintheta = r(3costheta - sintheta)r2cos2θ=3rcosθ−rsinθ=r(3cosθ−sinθ)

(divide both sides by r )

hence rcos^2theta =3 costheta - sintheta rcos2θ=3cosθ−sinθ

rArr r =( 3costheta -sintheta)/cos^2theta = 3costheta/cos^2theta - sintheta/cos^2theta ⇒r=3cosθ−sinθcos2θ=3cosθcos2θ−sinθcos2θ

= 3/costheta - tantheta/costheta = 3sectheta - tanthetasectheta=3cosθ−tanθcosθ=3secθ−tanθsecθ

rArr sectheta (3 - tantheta ) ⇒secθ(3−tanθ)

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