Eccentricity, trouble w/ the proof

I'm trying to understand the proof for eccentricity; I can get about halfway through it. The problem I'm having is summarized below

If e<1, you get the equation of an ellipse of the form

$\displaystyle \frac{(x-h)^2}{a^2}+\frac{y^2}{b^2}=1$

where

Eq. 4:

$\displaystyle h= -\frac{e^2d}{1-e^2}$

$\displaystyle a^2= -\frac{e^2d^2}{(1-e^2)^2}$

$\displaystyle b^2= -\frac{e^2d^2}{1-e^2}$

The foci of an ellipse are at a distance c from the center, where

Eq 5:

$\displaystyle c^2=a^2-b^2=$$\displaystyle \frac{e^4d^2}{(1-e^2)^2}$

This shows that

$\displaystyle c= \frac{e^2d}{1-e^2}=-h$

It follows from equations 4 and 5 that the eccentricity is given by

$\displaystyle e=\frac{c}{a}$

If someone could help explain how $\displaystyle e=\frac{c}{a}$ is derived, or really what it means I'd be so grateful. I'm using Calculus Early Transcendentals 6th Ed. by Stewart (Ch. 10, Sect. 6).

Re: Eccentricity, trouble w/ the proof

The conclusion is that the eccentricity of an ellips is: $\displaystyle e=\frac{c}{a}$, now:

If you write: $\displaystyle e^2=\frac{c^2}{a^2}$

Calculating with this:

$\displaystyle \frac{c^2}{a^2}=\frac{a^2-b^2}{a^2}=1-\frac{b^2}{a^2}$

Substitution of $\displaystyle b^2$ and $\displaystyle a^2$ out of eq 4 and you get:

$\displaystyle 1-\frac{b^2}{a^2}=1-\left[\frac{\frac{-e^2d^2}{1-e^2}}{\frac{-e^2d^2}{(1-e^2)^2}}}\right]$

$\displaystyle =1-(1-e^2)=e^2$

Now take the square root of $\displaystyle e^2=e$.