Find the Cartesian equation of the polar curve
$$r\cos2\theta = \frac1r+\cos\theta$$Options:
- $x^2-y^2=1+x$
- $x^2-y^2=1-x$
- $x^2+y^2=1+x$
- $x^2+y^2=1-x$
- None of the above.
I've solved both sides of the equation separately:
$$r\cos(2\theta) = r\cos^2(\theta) – r\sin^2(\theta) = x^2-y^2 \tag{1}$$
and
$$\frac1r+\cos(\theta) \quad\stackrel{\times r}{\to}\quad 1+r\cos(\theta) = 1+x \tag{2}$$
But I'm not sure if I can just throw the two sides of the equation together and call it a day? I don't think I can since if I multiply one side by $r$ then I must do it to the other side which messes up my answer.
Thanks in advance!
Best Answer
Multiplication by $r$ is the right way forward, but as @Blue notes$$x^2-y^2=r^2(\cos^2\theta-\sin^2\theta)=r^2\cos2\theta,$$so the problem you fear never arises. (Dimensional analysis is a good sanity check for problems like this, as it would have exposed your power miscounting.) So$$0=r^2\cos2\theta-r\cos\theta-1=x^2-y^2-x-1,$$which is option 1. As a geometric aside, this is a hyperbola, $(x-\tfrac12)^2-y^2=\tfrac54$.