Homework Help: Polar Conic to Cartesian

1. Aug 16, 2009

Knissp

1. The problem statement, all variables and given/known data
The equation of a conic in polar coordinates is:
$$r = \frac{r_o}{1-\epsilon cos(\theta)}$$.

$$\epsilon$$ is the eccentricity, 0 for a circle, (0,1) for an ellipse, 1 for a parabola, and >1 for a hyperbola.

What is this equation expressed in Cartesian coordinates?

2. Relevant equations
$$r=(x^2+y^2)^{1/2}$$
$$x=r cos(\theta)$$

3. The attempt at a solution
$$r = \frac{r_o}{1-\epsilon cos(\theta)}$$

$$r (1-\epsilon cos(\theta)) = r_o$$

$$r - r cos(\theta) \epsilon = r_o$$

$$r - x \epsilon = r_o$$

$$r = r_o + x \epsilon$$

$$(r)^2 = (r_o + x \epsilon)^2$$

$$x^2 + y^2 = r_o^2 + 2 (x \epsilon) r_o + (x \epsilon)^2$$

$$x^2 - (x \epsilon)^2 - 2 \epsilon r_o x + y^2 = r_o^2$$

$$(1 - \epsilon^2)x^2 - 2 \epsilon r_o x + y^2 = r_o^2$$

My textbook says this equation should be:

$$(1 - \epsilon^2)x^2 - 2 r_o x + y^2 = r_o^2$$

(Notice that we differ on the coefficient of x. Is the textbook missing an epsilon by a misprint? or did I mess up somewhere?

Thanks for the help.

2. Aug 16, 2009

tiny-tim

Hi Knissp!

(have an epsilon: ε )

No, you're correct …

if we put ε = 0, for a circle, then your equation is obviously correct, but the book's equation is (x - r0)2 + y2 = 2r02, which is a circle, but not the right one!