Elliptical Orbit and Kepler's equation

[DuckAmuck]

TL;DR

Trying to make sense of the math here.

It is easy to find that the equation for an ellipse is:
$$1 = x^2/a^2 + y^2/b^2$$
Then according to Kepler's equation:
$$x = a(\cos(E)-e)$$
$$y = b\sin(E)$$
where E is the eccentric anomaly and e is the eccentricity.

If you plug the Kepler's equations' x and y into the equation for the ellipse, you get a relationship that does not hold:
$$\cos(E) = e/2$$
Are the Kepler's equations approximate? I thought they were exact. What is wrong here?

 

[phyzguy]

Try reading the Wiki entry on Eccentric anomaly and studying the diagram. When you write:
\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1
You are implicitly assuming that the (0,0) point is at the center of the ellipse, not at the focus. In this coordinate system, x = a \cos(E), and y = b \sin(E). The formulas you gave for x and y are in the coordinate system where (0,0) is at the focus of the ellipse (i.e. at the sun). In this coordinate system:
\frac{(x + a e)^2}{a^2} + \frac{y^2}{b^2} = 1
Then everything works out. Note that the diagram in the Wiki page on Eccentric Anomaly is wrong. The distance from the center to the focus is a*e, not e.