Classical phase and space equal-a-priori postulate This question is vary related to the Phys.SE question here.

The objective of the problem is to find the classical probability of finding the position of a one-dimensional harmonic oscillator 

$$E = \frac{p^2}{2m} + \frac{kx^2}{2}$$

that it is given by ($A := \sqrt{2E/k}$)

$$P(x) = \frac{1}{2\pi A\sqrt{1 - x^2/A^2}}$$

We know that in phase space the micro-states acessible by a energy between $E$ and $E+\delta E$ is related to a area given by the elliptic shell 
If we name the total area of the elliptic shell by $\tilde{A}$ and $\delta \tilde{A}$ by the small elliptic shell between $x$ and $x + \mathrm d x$ for $x \in \left[\sqrt{2E/k},\sqrt{2(E+\delta E)/k}\right]$. The question is then 

Prove that it is possible to find $P(x)\mathrm dx$ using $\delta \tilde{A}/\tilde{A}$


My attempt: The area of an ellipse is given by $\pi a b$ where $a,b$ are the semi axis, for the problem, we have that $\tilde{A} = 2\pi\sqrt{m/k} \left(E + \delta E - E\right)  = 2 \pi \sqrt{m/k}\delta E $. And we could use the momentum to find the $\delta \tilde{A}$ where we can use that 
$$\delta \tilde{A} = \pi ((x + \mathrm d x)p(x + \mathrm dx) - xp(x)) = \pi x \frac{\mathrm d p}{\mathrm d x}\mathrm d x + \pi p(x + \mathrm d x)\mathrm d x $$
But how we can relate both to get the correct ratio and find what the problem asks?
We can use that the momentum $p$ can be wright as (using the equation for $E$ above)
$$p(x) = A\sqrt{mk}\sqrt{1 - x^2/A^2}$$
And insert in the relation for $\delta A$ (we can use $p(x + \mathrm d x) \approx p(x)$) adn this leads to the incorrect answer.

Physical Consept of the question: Using the above link and the answer here we prove for the harmonic oscillator the equivalence of the equal-a-priori postulate and the Ergodic Hypotesis. Reference [Problem 4; Chap. 2.] 
 A: The probability is given by $$P(x) dx = \frac{\mathcal{A}_1}{\mathcal{A}_2}$$
where $\mathcal{A}_1$ is the area of the two rectangular shapes between the two ellipses and the vertical lines $x$ and $x+dx$, and $\mathcal{A}_2$ is the full area between the two ellipses. We compute 
$$\mathcal{A}_2 = 2 \pi \, dE \,  \sqrt{m/k}$$
$$\mathcal{A}_1 = 2 \left[\sqrt{2m \left(E +dE-\frac{kx^2}{2} \right)} - \sqrt{2m \left(E -\frac{kx^2}{2} \right)} \right] dx \sim \frac{\sqrt{2m} \, dE \,  dx}{\sqrt{\left(E -\frac{kx^2}{2} \right)}} \, . $$ 
The $2$ in front of the bracket is here to count the two rectangular shapes. Then 
$$P(x)\, dx = \frac{\sqrt{2m} \, dE \,  dx}{2 \pi \, dE \,  \sqrt{m/k}\sqrt{E -\frac{kx^2}{2}}} = \frac{  dx}{ \pi \sqrt{2E/k} \sqrt{1 -\frac{kx^2}{2E}}}  = \frac{  dx}{ \pi A \sqrt{1 -\frac{x^2}{A^2} }}$$
The conclusion is that 
$$P(x)=\frac{1}{ \pi A \sqrt{1 -\frac{x^2}{A^2} }} \, . $$
You have a factor of 2 missing in your formula. Note that you can check that the above expression is correct by integrating, $\int_{-A}^A \frac{  dx}{ \pi A \sqrt{1 -\frac{x^2}{A^2} }} = 1$. 
