How do you convert #r = 8 + 5cos(theta)# into cartesian form?

1 Answer
John D.
Jul 20, 2017

#x^4 - 10x^3 - 39x^2+2x^2y^2-10xy^2-64y^2+y^4 = 0#

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# is always positive, so we don't need the #+-# sign in front of our square root when making that substitution

#r = 8 + 5costheta#

#r^2 = 8r +5rcostheta#

#x^2+y^2 = 8sqrt(x^2+y^2) + 5x#

Since we have #x# and #y# mixed together, we shouldn't try to separate them out and solve for #y# by itself. Instead, let's try to get this equation into the form #f(x,y) = 0#.

#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

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