1. ## Polar Coordinates

Hi. I need to figure out how to convert the polar equation
$\displaystyle r = 2cos \theta - sin \theta$
to cartesian coordinates and then graph the resulting equation.

I know it's an ellipse and I know $\displaystyle x = rcos \theta$, $\displaystyle y = rsin \theta$ and $\displaystyle r^2 = x^2 + y^2$, but I can't quite figure out how to make this equation into the standard form for an ellipse. The closest I've come is $\displaystyle \frac{x^2-2x}{2} + \frac{y^2 + y}{2} = 0$, which doesn't make any sense. Could someone help me? (Also, I wasn't sure whether this belonged in the calculus or geometry forum, but it's for my calc class so I posted it here.)

2. First, multiply both sides of your polar equation by r:
$\displaystyle r^2=2r\cos\theta-r\sin\theta$
Now, use $\displaystyle x=r\cos\theta$, $\displaystyle y=r\sin\theta$, and $\displaystyle r^2=x^2+y^2$, like you said, to get:
$\displaystyle x^2+y^2=2x-y$
Rearranging:
$\displaystyle x^2-2x+y^2+y=0$
Completing the square on both the x and y terms:
$\displaystyle x^2-2x+1+y^2+y+\frac{1}{4}=\frac{5}{4}$
$\displaystyle (x-1)^2+\left(y+\frac{1}{2}\right)^2=\frac{5}{4}$
You should be able to recognise what sort of conic section this is, and be able to graph it.

--Kevin C.

3. Originally Posted by TwistedOne151
First, multiply both sides of your polar equation by r:
$\displaystyle r^2=2r\cos\theta-r\sin\theta$
Now, use $\displaystyle x=r\cos\theta$, $\displaystyle y=r\sin\theta$, and $\displaystyle r^2=x^2+y^2$, like you said, to get:
$\displaystyle x^2+y^2=2x-y$
Rearranging:
$\displaystyle x^2-2x+y^2+y=0$
Completing the square on both the x and y terms:
$\displaystyle x^2-2x+1+y^2+y+\frac{1}{4}=\frac{5}{4}$
$\displaystyle (x-1)^2+\left(y+\frac{1}{2}\right)^2=\frac{5}{4}$
You should be able to recognise what sort of conic section this is, and be able to graph it.

--Kevin C.
Thank you Kevin. I always forget to complete the square.