# How to translate from polar to cartesian coordinates:

1. Jun 5, 2013

### sjombol

How to translate r = 2 /(2 - cos(theta)) to cartesian coordinates:

so far:

r = 2 /(2 - cos(theta))

r = 2 /(2 - cos(theta)) |* (2 - cos(theta)) both sides

r (2 - cos(theta))= 2

2*r - rcos(theta) = 2 | know x = rcos(theta)

2*r - x = 2 | know r^2 = x^2 + y^2

2*(x^2 + y^2)^1/2 - x = 2 | : 2

(x^2 + y^2)^1/2 - x/2 = 1 | opens the parentheses

x^2*1/2 + y^2*1/2 - x/2 = 1

x + y - x/2 = 1

y = 1 - x/2 | is this the solution?

Thank you in advance for help.

Regards
sjombol

2. Jun 5, 2013

### LCKurtz

Is $\sqrt{a+b} = \sqrt a + \sqrt b$?

Last edited: Jun 5, 2013
3. Jun 5, 2013

### Staff: Mentor

It's interesting that you document each step along the way with what you did, and I think that's a good thing. Where you fell down, as LCKurtz notes, is in the step that you describe as "opens the parentheses". The distributive property, which is usually represented something like this -- a(b + c) = ab + ac -- might be what you were thinking. The problem is that that this property applies only to multiplication of a sum. It doesn't apply to expressions raised to powers (including square roots) or other operations.

4. Jun 5, 2013

### sjombol

It's been a while since last time I went to school, guess old habits do die sometimes or at least needs refreshing.

Another go then:

r = 2 /(2 - cos(theta))

r = 2 /(2 - cos(theta)) |* (2 - cos(theta)) both sides

r (2 - cos(theta))= 2

2*r - rcos(theta) = 2 | know x = rcos(theta)

2*r - x = 2 | know r^2 = x^2 + y^2

2*(x^2 + y^2)^1/2 - x = 2 | move x to right side and both sides ^2, gives me:

2^2 * ((x^2 + y^2)^1/2)^2 = (x + 2)^2

4 * (x^2 + y^2) = (x + 2)^2

4x^2 + 4y^2 = x^2 + 4x + 4 | move 4x^2 to right side

4y^2 = x^2 + 4x + 4 - 4x^2

4y^2 = -3x^2 + 4x + 4 | divide by 4 and takes the square root, giving me:

y = ((-3x^2 + 4x + 4)/4)^1/2 | the calculator draws this as an half ellipse, above x-axis ?

Last edited: Jun 5, 2013
5. Jun 5, 2013

### LCKurtz

I didn't check all your steps, but if you only took the positive square root, it wouldn't surprise me if half the graph is missing.

6. Jun 6, 2013

### HallsofIvy

Staff Emeritus
Why take the square root? You understand that you will be losing half the graph that way, don't you?