<s>[INST] Correlation coefficient for a uniform distribution on an ellipse I am currently reading a paper that claims that the correlation coefficient for a uniform distribution on the interior of an ellipse
$$f_{X,Y} (x,y) = \begin{cases}\text{constant} & \text{if} \ (x,y) \ \text{inside the ellipse} \\  0 & \text{otherwise}  \end{cases}$$
is given by
$$\rho = \sqrt{1- \left(\frac{h}{H}\right)^2 }$$
where $h$ and $H$ are the vertical heights at the center and at the extremes respectively.

The author does not reveal how he reaches that and instead only says that we need to change scales, rotate, translate and of course integrate. I would very much like to retrace his steps but I am a bit lost with all that. I would therefore be grateful for some hints.
Thank you in advance.
Oh and for the record

Châtillon, Guy. "The balloon rules for a rough estimate of the correlation coefficient." The American Statistician 38.1 (1984): 58-60

It's quite amusing.
 [/INST] Let $(X,Y)$ be uniformly distributed on the interior of the ellipse
$$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$ where $a$ and $b$ are the
semi-axes of the ellipse.  Then, $X$ and $Y$ have marginal densities
\begin{align}
f_X(x) &= \frac{2}{\pi a^2}\sqrt{a^2-x^2}\,\,\mathbf 1_{-a,a}(x),\\
f_X(x) &= \frac{2}{\pi b^2}\sqrt{b^2-y^2}\,\,\mathbf 1_{-b,b}(y),
\end{align}
and it is easy to see that $E[X] = E[Y] = 0$.  Also,
\begin{align}
\sigma_X^2 = E[X^2] &= \frac{2}{\pi a^2}\int_a^a x^2\sqrt{a^2-x^2}\,\mathrm dx\\
&= \frac{4}{\pi a^2}\int_0^a x^2\sqrt{a^2-x^2}\,\mathrm dx\\
&= \frac{4}{\pi a^2}\times a^4 \frac 12\frac{\Gamma(3/2)\Gamma(3/2)}{\Gamma(3)}\\
&= \frac{a^2}{4},
\end{align}
and similarly, $\sigma_Y^2 = \frac{b^2}{4}$. Finally, 
$X$ and $Y$ are uncorrelated random variables.
Let \begin{align}
U &= X\cos \theta - Y \sin \theta\\
V &= X\sin \theta + Y \cos \theta
\end{align}
which is a rotation transformation applied to $(X,Y)$. Then,
$(U,V)$ are uniformly distributed on the interior of an
ellipse whose axes do not coincide with the $u$ and $v$ axes.
But, it is easy to verify that $U$ and $V$ are zero-mean
random variables and that their variances are 
\begin{align}
\sigma_U^2 &= \frac{a^2\cos^2\theta + b^2\sin^2\theta}{4}\\
\sigma_V^2 &= \frac{a^2\sin^2\theta + b^2\cos^2\theta}{4}
\end{align}
Furthermore, 
$$\operatorname{cov}(U,V) = (\sigma_X^2-\sigma_Y^2)\sin\theta\cos\theta
= \frac{a^2-b^2}{8}\sin 2\theta$$
from which we can get the value of $\rho_{U,V}$.
Now, the ellipse on whose interior $(U,V)$ is uniformly distributed
has equation
$$\frac{(u \cos\theta + v\sin \theta)^2}{a^2}
+ \frac{(-u \sin\theta + v\cos \theta)^2}{b^2} = 1,$$
that is,
$$\left(\frac{\cos^2\theta}{a^2} + \frac{\sin^2\theta}{b^2}\right) u^2  + \left(\frac{\sin^2\theta}{a^2} + \frac{\cos^2\theta}{b^2}\right) v^2 
+ \left(\left(\frac{1}{a^2} - \frac{1}{b^2}\right)\sin 2\theta \right)uv = 1,$$
which can also be expressed as
$$\sigma_V^2\cdot u^2 + \sigma_U^2\cdot v^2 
-2\rho_{U,V}\sigma_U\sigma_V\cdot uv = \frac{a^2b^2}{4}\tag{1}$$
Setting $u = 0$ in $(1)$ gives 
$\displaystyle h  = \frac{ab}{\sigma_U}$.
while implicit differentiation of $(1)$ with respect to $u$ gives
$$\sigma_V^2\cdot 2u + \sigma_U^2\cdot 2v\frac{\mathrm dv}{\mathrm du} 
-2\rho_{U,V}\sigma_U\sigma_V\cdot 
\left(v + u\frac{\mathrm dv}{\mathrm du}\right) = 0,$$
that is, the tangent to the ellipse $(1)$ is horizontal at
the two points $(u,v)$ on the ellipse for which
$$\rho_{U,V}\sigma_U\cdot v = \sigma_v\cdot u.$$
The value of $H$ can be figured out from this, and will (in the
unlikely event that I have made no mistakes in doing the
above calculations) lead to the desired result.
</s>