Convert $y^2 = 4(x + 1)$ to a polar equation I'm trying to convert the rectangular cartesian equation

$$
y^2 = 4(x + 1)
$$

to a polar equation. After replacing $y = r \sin \theta$ and $x = r \cos \theta$, I get

$$
r^2 \sin^2 \theta = 4(r \cos \theta + 1)
$$

After replacing $\sin^2 \theta = 1 - \cos^2 \theta$ and rearranging, I get

$$
r^2 - r^2 \cos^2 \theta - 4r \cos \theta -4 = 0
$$

That's where I'm stuck and I can't solve the equation in terms of $r$.
Basic approach. Rewrite your equation as

$$
(\sin^2\theta) r^2 - (4\cos\theta) r - 4 = 0
$$

and use the quadratic formula for $r$.
$$r^2−r^2\cos^2θ−4r\cosθ−4=0$$

$$\implies r^2 = r^2\cos^2θ+4r\cosθ+4$$

If  $r\cosθ = P$
, then

$$r^2 = P^2 + 4P + 4$$

$$\implies r^2 = (P+2)^2$$

$$\implies r = P + 2$$Now

$$r = r\cosθ + 2$$

$$\implies r - r\cosθ = 2$$

$$\implies r(1-\cosθ) = 2$$

$$\implies r = 2/(1-\cosθ)$$