How Do You Convert a Polar Conic Equation to Cartesian Coordinates?

[Knissp]

Homework Statement

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?

Homework Equations

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

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.

 

[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!

 

[Architeuthis Dux]

I would say that your approach is correct and the textbook may have a typo. The coefficient of x should indeed be 2 \epsilon r_o, as you have calculated. The equation provided by the textbook would not yield the correct graph for a conic in Cartesian coordinates. It is always important to double check formulas and equations, especially when they are provided in textbooks.