How do you convert this polar equation to Cartesian equation?

Question

I have this equation to solve:

$$r(1+\cos \theta) = 2$$

I know the answer is $y^2 = 4 - 4x$ but I don’t understand how to get there. I’ve tried multiplying both sides by $r$ as well as multiplying both sides by cos theta.

Answer 1

We have that

$$r(1+\cos \theta)=r+r\cos \theta=\sqrt{x^2+y^2}+x=2$$

and for $2-x\ge 0$

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

$$\iff x^2+y^2=4-4x+x^2 \iff y^2=4-4x$$

Answer 2

From your equation, we get

$$r=\frac{2}{1+\cos(t)}$$

$$x=r\cos(t)=\frac{2\cos(t)}{1+\cos(t)}$$ $$=\frac{2(2\cos^2(\frac t2)-1)}{2\cos^2(\frac t2)}$$ $$=2-\frac{1}{\cos^2(\frac t2)}$$ and $$y=r\sin(t)=\frac{2\sin(t)}{1+\cos(t)}$$

$$=\frac{4\sin(\frac t2)\cos(\frac t2)}{2\cos^2(\frac t2)}$$ $$=2\tan(\frac t2)$$

From here, we see that, using the well known formula $$1+\tan^2(X)=\frac{1}{\cos^2(X)}$$

$$1+\frac{y^2}{4}=2-x$$

and $$y^2=8-4x-4=4-4x$$

Answer 3

well, we always have $x = r\cos \theta$ and $y=r\sin\theta$; always. And we always have $r = \sqrt{x^2 + y^2}$ and $\cos \theta = \frac {x}{\sqrt{x^2 + y^2}}$ and $\sin \theta =\frac {y}{\sqrt{x^2 + y^2}}$ (and so $\theta = \arctan \frac yx$). Always.

So this equation is $\sqrt{x^2 + y^2}(1 + \frac x{\sqrt{x^2 + y^2}}) = 2$. Some basic algebraic manipulation

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

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

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

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

$y^2 = 4-4x$.