How do you convert #r = 1/(8 - cos(theta)) # into rectangular form?

1 Answer
Alan P.
Dec 7, 2015

#8sqrt(x^2+y^2)-x=1#

Explanation:

In converting between polar and rectangular forms:
#color(white)("XXX")r=sqrt(x^2+y^2)#
and
#color(white)("XXX")cos(theta)=x/r = x/(sqrt(x^2+y^2))#

Therefore, given
#color(white)("XXX")r = 1/(8-cos(theta))#

We have
#color(white)("XXX")r = 1/(8-x/r)#

#color(white)("XXX")r= 1/((8r-x)/r)#

#color(white)("XXX")r= r/(8r-x)#

#color(white)("XXX")8r-x=1#

#color(white)("XXX")8(sqrt(x^2+y^2))-x = 1#

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