How do you convert #r = 8 + 5cos(theta)# into cartesian form?
1 Answer
Explanation:
Remember the conversions:
#x = rcostheta#
#y = rsintheta#
#x^2+y^2=r^2#
#+-sqrt(x^2+y^2)=r#
Note that
#r = 8 + 5costheta#
#r^2 = 8r +5rcostheta#
#x^2+y^2 = 8sqrt(x^2+y^2) + 5x#
Since we have
#x^2-5x+y^2 = 8sqrt(x^2+y^2)#
#(x^2-5x+y^2)^2 = 64(x^2+y^2)#
#x^4-10x^3+2x^2y^2+25x^2-10xy^2+y^4 = 64x^2+64y^2#
#x^4 - 10x^3 - 39x^2+2x^2y^2-10xy^2-64y^2+y^4 = 0#
We could try to simplify this in other ways, but this is a valid form of this equation, so we'll leave it like this.
Final Answer