Calculus- Converting equation from Cartesian to polar Cordinates

Question

I'm having difficulties with the following equation to simplify it and converting to a polar form

$$y^2 - 3x^2 - 4x - 1 = 0$$

Answer 1

Put

$$\begin{cases}x=r\cos\theta\\{}\\y=r\sin\theta\end{cases}\implies y^2-3x^2-4x-1=0\;\mapsto r^2(\sin^2\theta-3\cos^2\theta)-4r\cos\theta-1=0$$

Another way:

$$0=y^2-3x^2-4x-1=y^2+x^2-4(x^2+x)-1\implies$$

$$x^2+y^2=4\left(x+\frac12\right)^2\stackrel{\text{polar coord.}}\implies r^2=4\left(r\cos\theta+\frac12\right)^2=4\frac{2r\cos\theta+1}2=4r\cos\theta+2$$