Proof for Axes of an Ellipse - Analytical Geometry

[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.

 

[QuantumQuest]

The polar equation of an ellipse is $r = \frac{p}{1 + e\cos \theta}$, where $p$ is the semi-latus rectum and $e$ is the eccentricity of the ellipse. The origin is a focus e.g. $F$ of the ellipse. The other focus of the ellipse is $F'$. When $\theta = 0^0$ we have $cos \theta = 1$ and for $\theta = 180^0$, $\cos \theta = -1$. So, if $a$ is the semi-major axis, we have:
$2a = \frac{p}{1 + e} + \frac{p}{1 - e} = \frac{2p}{1 - e^2}$. Hence, $a = \frac{p}{1 - e^2}$.

Now, let $c$ be the distance from the center (coordinate axes intersection) to either focus, let's say $F$.
We see that $c = a - \frac{p}{1 + e} = \frac{ep}{1 - e^2} = ea$. If $b$ is the semi-minor axis - e.g. in the upper half of the ellipse in the above diagram, and if we draw a line that connects the focus $F$ to the point that semi-minor axis $b$ intersects the ellipse, we have a right triangle. Its two perpendicular sides are $c$ and $b$ and its hypotenuse is $a$. Using Pythagorean theorem and utilizing the fact that $c = ea$, as we saw above, we find that $b = a \sqrt{1 - e^2}$. Now, using the expression we obtained before for $a$ we have

$b = \frac{p}{1 - e^2} \cdot \sqrt{1 - e^2}$

So, the (semi)axes $a$ and $b$ can indeed be written in a way, such that, if we square the final expressions we found, we have the expressions asked in the OP.