How do you change, this polar form equation #r(2 + cos theta) = 1# to Cartesian form?

1 Answer
Yonas Yohannes
Feb 15, 2016

#2sqrt(x^2+y^2) + x - 1 # I would leave it here optionally you can write (explanation)...
#5x^2+4y^2 + 4xsqrt(x^2+y^2) - 1 = 0 #

Explanation:

1) GIVEN: #r(2+costheta) = 1; r(theta) = 1/(2+cos(theta) #
Transform #r(theta) => f(x,y)# Polar => Cartesian
2) RECIPE: #r=x^2+y^2; y=rsintheta; x=rcostheta; theta = tan^-1(y/x)#
3) Solution:
a) Multiply by r(theta) by r:
#2r+rcostheta = 1#
b) Substitute from 2)
#2sqrt(x^2+y^2) + x = 1 # Now if you like you can square both sides or stop here...
#a^2 + 2ax + x^2 = 1; " where " a = 2sqrt(x^2+y^2) #
#4(x^2+y^2) + 4xsqrt(x^2+y^2) + x^2 = 1 #
#5x^2+4y^2 + 4xsqrt(x^2+y^2) - 1 = 0 #

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