How do you convert 8=(6x-4y)^2+2y-x8=(6x−4y)2+2y−x into polar form?

1 Answer
Shwetank Mauria
Jan 1, 2018

r^2(10cos2theta-24sin2theta+6)+r(2sintheta-costheta)-8=0r2(10cos2θ−24sin2θ+6)+r(2sinθ−cosθ)−8=0

Explanation:

The relation between polar coordinates (r,theta)(r,θ) and Cartesian coordinates (x,y)(x,y) is x=rcosthetax=rcosθ, y=rsinthetay=rsinθ and x^2+y^2=r^2x2+y2=r2.

Hence 8=(6x-4y)^2+2y-x8=(6x−4y)2+2y−x can be written as

8=36x^2+16y^2-48xy+2y-x8=36x2+16y2−48xy+2y−x

or 8=36r^2cos^2theta+16r^2sin^2theta-48r^2sinthetacostheta+2rsintheta-rcostheta8=36r2cos2θ+16r2sin2θ−48r2sinθcosθ+2rsinθ−rcosθ

or 8=20r^2cos^2theta+16r^2-24r^2sin2theta+2rsintheta-rcostheta8=20r2cos2θ+16r2−24r2sin2θ+2rsinθ−rcosθ

or r^2(20cos^2theta-24sin2theta+16)+r(2sintheta-costheta)-8=0r2(20cos2θ−24sin2θ+16)+r(2sinθ−cosθ)−8=0

or r^2(10(cos2theta-1)-24sin2theta+16)+r(2sintheta-costheta)-8=0r2(10(cos2θ−1)−24sin2θ+16)+r(2sinθ−cosθ)−8=0

or r^2(10cos2theta-24sin2theta+6)+r(2sintheta-costheta)-8=0r2(10cos2θ−24sin2θ+6)+r(2sinθ−cosθ)−8=0

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