1. ## Polar Coordinate Problem

Hello, I'm having a problem involving polar coordinates and writing an equation using rectangular coordinates. Here are the directions and the problem:

The letters r and θ represent polar coordinates. Write each equation using rectangular coordinates (x,y).

r=(3)/[3-cos(θ)]

I attempted to work on the problem, but I eventually got stuck. Here is my attempt:

r = (3)/[3-cos(θ)]
r[3-cos(θ)] = 3
3r - rcos(θ) = 3
[3r - rcos(θ)]2 = 32
9r2 - 6r2cos(θ) + r2cos(θ) = 9
9r2 - 5r2cos(θ) = 9

After that, I'm not exactly sure on what I need to do. I would turn the r2 into x2 + y2, but I'm not sure what to do with the 5r2cos(θ).

2. ## Re: Polar Coordinate Problem

Originally Posted by Smugleaf
Hello, I'm having a problem involving polar coordinates and writing an equation using rectangular coordinates. Here are the directions and the problem:

The letters r and θ represent polar coordinates. Write each equation using rectangular coordinates (x,y).

r=(3)/[3-cos(θ)]

I attempted to work on the problem, but I eventually got stuck. Here is my attempt:

r = (3)/[3-cos(θ)]
r[3-cos(θ)] = 3
3r - rcos(θ) = 3
[3r - rcos(θ)]2 = 32
9r2 - 6r2cos(θ) + r2cos(θ) = 9
9r2 - 5r2cos(θ) = 9

After that, I'm not exactly sure on what I need to do. I would turn the r2 into x2 + y2, but I'm not sure what to do with the 5r2cos(θ).

the coordinate transform equations are

$\displaystyle x = r \cos(\theta)$

$\displaystyle y = r \sin(\theta)$

so

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

LaTex interpretation is being dysfunctional tonight sorry.

3. ## Re: Polar Coordinate Problem

Hello, Smugleaf!

You squared prematurely . . .

$\displaystyle \text{Write in rectangular coordinates: }\:r\:=\:\frac{3}{3-\cos\theta}$

$\displaystyle r(3-\cos\theta) \:=\:3$

$\displaystyle 3r - r\cos\theta \:=\:3$

$\displaystyle 3\sqrt{x^2+y^2} - x \:=\:3$

$\displaystyle 3\sqrt{x^2+y^2} \:=\:x+3$

$\displaystyle \text{Square: }\:9(x^2+y^2) \:=\:x^2 + 6x + 9$

. . . . . . $\displaystyle 9x^2 + 9y^2 \:=\:x^2+6x+9$

. . $\displaystyle 8x^2 - 6x + 9y^2 \:=\:9$