How do you convert 1=(x+4)^2+(7y-2)^21=(x+4)2+(7y−2)2 into polar form?

1 Answer
Shwetank Mauria
Oct 4, 2016

r^2(1+48sin^2theta)+r(8costheta-28sintheta)+19=0r2(1+48sin2θ)+r(8cosθ−28sinθ)+19=0

Explanation:

The relation between rectangular Cartesian coordinates (x,y)(x,y) and polar coordinates (r,theta)(r,θ) is given by

x=rcosthetax=rcosθ, y=rsinthetay=rsinθ and x^2+y^2=r^2x2+y2=r2

Hence y^2+(x-3)^2=9y2+(x−3)2=9 can be rewritten as

(x+4)^2+(7y-2)^2=1(x+4)2+(7y−2)2=1

or (rcostheta+4)^2+(7rsintheta-2)^2=1(rcosθ+4)2+(7rsinθ−2)2=1

or (r^2cos^2theta+8rcostheta+16)+(49r^2sin^2theta-28rsintheta+4)=1(r2cos2θ+8rcosθ+16)+(49r2sin2θ−28rsinθ+4)=1

or r^2cos^2theta+8rcostheta+49r^2sin^2theta-28rsintheta+20-1=0r2cos2θ+8rcosθ+49r2sin2θ−28rsinθ+20−1=0

or r^2(1+48sin^2theta)+r(8costheta-28sintheta)+19=0r2(1+48sin2θ)+r(8cosθ−28sinθ)+19=0

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