# Axes of an ellipse. I need help on this!

1. Mar 20, 2007

### sutupidmath

in my textbook says that a^2=p^2/(1-e^2)^2, and b^2=p^2/(1-e^2), are two axes of an ellipse, however there is no any proof as to how we can be sure that a and b are such axes. Where p is the focal parameter, and e is the eccentricity of the ellipse; a- is the big semi-axes, b- the small one.So i would like to know is there any proof that ensures us that the above expressions are indeed or represent the axes of any ellipse??
I asked the assistant proffesor on my Analytical Geometry class, but she did not know how to proof it.

Any help would be welcomed.

sorry for reposting it.

2. Mar 21, 2007

### HallsofIvy

Staff Emeritus
I wonder if your professor wasn't just hoping you would think about it more yourself- it's a simple calculation.

For any ellipse, the distance, d, from one focus to the ellipse and back to the other focus is a constant.

For an ellipse with major axis a along the x-axis, minor axis b, equation
$$\frac{x^2}{a^2}+ \frac{y^2}{b^2}= 1[/itex] the total distance from one focus, to the point (a, 0) to the other focus is d= 2a (that should be obvious- the distance you don't cover, from a focus to (-a, 0), is exactly the distance you cover twice, from the other focus to (a, 0)). Now, taking c to be the distance from the center to a focus (so the foci are at (c, 0) and (-c, 0)), the focal distance, we have, by the Pythagorean theorem, that (d/2)2= a2= b2+ c2 so that c2= a2- b2. The "eccentricity" is defined to be the ratio of focal distance to the length of the major-semiaxis, here that is [tex]e= \frac{\sqrt{a^2- b^2}}{a}= \sqrt{1- \frac{b^2}{a^2}}$$

The focal parameter, p, is the distance from the focus to the nearest directrix. For an ellipse, a directrix is a line perpendicular to the major axis such that the ratio of the distance from any point on to the nearest focus to the distance from that point to the nearest directrix is equal to the eccentricity. That is, with x the x coordinate of any point on the directix,
[tex]\frac{x-a}{a-c}= e[/itex]
Since c= ae, that is
[tex]\frac{x- a}{a- ae}= \frac{x- a}{a(1-e)}= e[/itex]
so that x- a= ae(1-e) and x= a+ ae(1-e)= a(1+ e- e^2).
Since x is the distance from the center of the ellipse to the directrix, p, the distance from the focus to the directrix is p= x- c= x- ae= a(1+ e- e^2)- ae= a(1- e^2). From that, a= p/(1- e2) which is equivalent to a2= p2/(1- e2)2, of course.

Since b2= a2- c2= a2- a2e2= a2(1- e2), we have b2= (p2/(1- e2)2)(1-e2)= p2/(1- e2).

3. Mar 22, 2007

### sutupidmath

thnx
It's all clear to me now.