My textbook says the polar equation, $r=1+2r\cos \theta$, its cartesian equivalent is $y^2-3x^2-4x-1=0.$ I understand that I get this if I square $r$; $r^2=x^2+y^2=(1+2x)^2.$ But don't I need to square root this back or something?? I thought the answer is supposed to be $ \sqrt{y^2-3x-4x-1}=0 $… I'm confused..
[Math] changing $r=1+2r\cos \theta$ to its cartesian equivalent
calculuspolar coordinates
Best Answer
From $r=1+2r\cos\theta$, we get $$r(1-2\cos\theta)=1$$ or $$r(1-2\frac xr)=1$$ $$r-2x=1$$ $$\sqrt{x^2+y^2}=2x+1$$ and finally square to get $$x^2+y^2=4x^2+4x+1$$