# Eccentricity, trouble w/ the proof

• Jul 9th 2011, 11:54 AM
anything
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

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

where

Eq. 4:
$h= -\frac{e^2d}{1-e^2}$
$a^2= -\frac{e^2d^2}{(1-e^2)^2}$
$b^2= -\frac{e^2d^2}{1-e^2}$

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

Eq 5:
$c^2=a^2-b^2=$ $\frac{e^4d^2}{(1-e^2)^2}$

This shows that

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

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

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

If someone could help explain how $e=\frac{c}{a}$ 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).
• Jul 9th 2011, 01:46 PM
Siron
Re: Eccentricity, trouble w/ the proof
The conclusion is that the eccentricity of an ellips is: $e=\frac{c}{a}$, now:
If you write: $e^2=\frac{c^2}{a^2}$
Calculating with this:
$\frac{c^2}{a^2}=\frac{a^2-b^2}{a^2}=1-\frac{b^2}{a^2}$
Substitution of $b^2$ and $a^2$ out of eq 4 and you get:
$1-\frac{b^2}{a^2}=1-\left[\frac{\frac{-e^2d^2}{1-e^2}}{\frac{-e^2d^2}{(1-e^2)^2}}}\right]$
$=1-(1-e^2)=e^2$
Now take the square root of $e^2=e$.