How do you convert #r^(2)=cos(θ)# to rectangular form?

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Answer 1 · 2016-05-28T06:04:57

#x^6+3x^4y^2+3x^2y^4+y^6-x^2=0, x>=0#.

#r^2=cos theta>=0 and <=1#.

So, theta is in the 1st quadrant or in the 4th, and

therefore, #x>=0 and <=1#..

#(r cos theta, r sin theta )=(x, y). r=sqrt(x^2+y^2) and cos theta=x/r#.

So, #r^2=x^2+y^2=x/sqrt(x^2+y^2)#. Rationalizing,

#(x^2+y^2)^3=x^6+3x^4y^2+3x^2y^4+y^6=x^2, x>=0 and <=1#.

The graph passes through (0, 0) and (1, 0). It is like an oval that is

symmetrical about the x-axis