Converting Polar Equation to Cartesian Equation

Question

Heading ##Convert polar equation to Cartesian equation.

$$r= \frac{2}{1-\cos\theta}$$

I tried to answer this and this is how I answered it. Please review if it's correct or not. Thank you! :)

\begin{align*} r\cos\theta & = x\\ r\sin\theta & = y\\ r & = \sqrt{x^2+y^2} \end{align*}

I multiplied both sides by $1- \cos\theta$. $$r(1-\cos\theta) = \frac{2}{1-\cos\theta} \cdot (1-\cos\theta)$$

Then, it will become \begin{align*} r(1-\cos\theta) & = 2\\ r - r\cos\theta & = 2\\ r - x & = 2\\ r & = x + 2\\ \sqrt{x^2+y^2} & = x+2 \end{align*}

Then, I multiplied both sides with 2 to eliminate the root. \begin{align*} 2\sqrt{x^2+y^2} & = 2(x+2)\\ x^2+y^2 & = (x+2)^2\\ x^2+y^2 & = x^2+4x+4 \end{align*}

Combine like terms. \begin{align*} x^2-x^2+y^2-4x+4 & = 0\\ y^2- 4x - 4 & = 0\\ y^2 & = 4x+4 \end{align*}

Answer 1

$$r-r\cos\theta=2\iff r=r\cos\theta+2$$

$$x+2=\sqrt{x^2+y^2}$$

and $$x^2+y^2=(x+2)^2\iff4x+4=y^2$$

Answer 2

From the diagram, you should be able to deduce the following relationships. $$\tan\theta = \frac{y}{x}$$ $$x = r\cos\theta$$ $$y = r\sin\theta$$ $$r^2=x^2+y^2$$ Now, try to solve the problem using stategic substitutions.