Exponential moving average cut-off frequency
The discrete recurrence relation given on the linked page is
$y[n] = (1-\alpha)y[n-1] + \alpha x[n]$
with $x[n]$ being the input samples and $y[n]$ being the output samples.
So taking the Z-transform
$Y(z) = (1-\alpha)z^{-1}Y(z) + \alpha X(z)$
$H(z) = \dfrac{Y(z)}{X(z)}= \dfrac{\alpha}{1-(1-\alpha)z^{-1}}$
To find the -3 dB corner of the filter, in terms of the normalized digital radian frequency, $\Omega$ in the range $[0, \pi]$, one solves
$\left|H(z=e^{j\Omega_{3dB}})\right|^2 = \dfrac{1}{2} = \left|\dfrac{\alpha}{1-(1-\alpha)e^{-j\Omega_{3dB}}}\right|^2$
$ = \dfrac{\alpha^2}{\left|1-(1-\alpha)\cos(-\Omega_{3dB}) -j (1-\alpha)\sin(-\Omega_{3dB})\right|^2}$
which results in this equation
$\left[1-(1-\alpha)\cos(\Omega_{3dB})\right]^2+\left[(1-\alpha)\sin(\Omega_{3dB})\right]^2 = 2\alpha^2$
After some more algebra and trigonometry, I come up with
$\Omega_{3dB} = \cos^{-1}\left[1-\dfrac{\alpha^2}{2(1-\alpha)}\right]$
and since
$f_{3dB} = \dfrac{\Omega_{3dB}}{\pi} \cdot \dfrac{F_s}{2}$
I come up with
$f_{3dB} = \dfrac{F_s}{2\pi}\cos^{-1}\left[1-\dfrac{\alpha^2}{2(1-\alpha)}\right]$
Here's some Octave/MatLab code so you can verify the result on a plot; just change your Fs and alpha as required:
Fs = 10000.0
alpha = 0.01

b = [alpha];
a = [1 -(1-alpha)];

[H, W] = freqz(b, a, 1024, Fs);

f3db = Fs/(2*pi)*acos(1-(alpha^2)/(2*(1-alpha)))

plot(W, 20*log10(abs(H)), [f3db, f3db], [-40, 0]);
grid on

Update in response to comment:
The graph is a plot of the filter's magnitude response in dB (on the y-axis) vs. input frequency in Hz (on the x-axis).  The vertical line designates the -3 dB corner frequency.
Since you want to specify your -3 dB corner frequency to find the value for $\alpha$, let's start with this equation from above:
$\left[1-(1-\alpha)\cos(\Omega_{3dB})\right]^2+\left[(1-\alpha)\sin(\Omega_{3dB})\right]^2 = 2\alpha^2$
and after some algebra and trigonometry, one can get a quadratic equation in $\alpha$
$\alpha^2 +2(1- \cos(\Omega_{3dB}))\alpha - 2(1- \cos(\Omega_{3dB})) = 0$
which has solutions
$\alpha = \cos(\Omega_{3db}) - 1 \pm \sqrt{\cos^2(\Omega_{3dB}) -4\cos(\Omega_{3dB}) +3}$
of which only the $+$ answer from the $\pm$ can yield a positive answer for $\alpha$.
Using the above solution for $\alpha$ and the relation
$\Omega_{3dB} = \dfrac{\pi}{F_s/2}f_{3dB}$
one can find $\alpha$ given $f_{3dB}$ and $F_s$.
Here is an updated Octave script:
Fs = 40000
f3db = 1

format long;
omega3db = f3db * pi/(Fs/2)

alpha = cos(omega3db) - 1 + sqrt(cos(omega3db).^2 - 4*cos(omega3db) + 3)

b = [alpha];
a = [1 -(1-alpha)];

[H, W] = freqz(b, a, 32768, Fs);

figure(1);
plot(W, 20*log10(abs(H)), [f3db, f3db], [-40, 0]);
xlabel('Frequency (Hz)');
ylabel('Magnitude Response (dB)');
title('EWMA Filter Frequency Response');
grid on;


W2 = [0:75] * pi/(Fs/2); % 0 to 75 Hz
H2 = freqz(b, a, W2);
W2 = W2 / (pi/(Fs/2));

figure(2);
plot(W2, 20*log10(abs(H2)), [f3db, f3db], [-20, 0]);
xlabel('Frequency (Hz)');
ylabel('Magnitude Response (dB)');
title('EWMA Filter Frequency Response near DC');
grid on;

Additional information in response to Sep 11 '20 comment
For a stable, low pass filter, the pole, $(1-\alpha)$, must be inside the unit circle of the z-plane and be greater than $0$ on the real axis.  So we must satisfy
$$0 &lt; 1 - \alpha &lt; 1$$
thus
$$0 &lt; \alpha &lt; 1$$
These are bounds on $\alpha$ (and not a &quot;definition&quot;) that come from requirements for stability and a filter passband centered at DC.
However, not all valid values of $\alpha$ will yield a -3 dB frequency that exists in the interval from DC to the Nyquist frequency, $\Omega \in [0, \pi]$ or equivalently $f \in \left[0, \dfrac{F_s}{2}\right]$.
When the pole, $(1-\alpha)$, is very close to the origin of the z-plane, the filter response will be so gradual or flat, that it will never drop to -3 dB or below.  Specifically, when
$$\Omega_{3dB} = \cos^{-1}\left[1-\dfrac{\alpha^2}{2(1-\alpha)}\right]$$
is undefined, due to
$$\left|1-\dfrac{\alpha^2}{2(1-\alpha)}\right| &gt; 1$$
the filter response is so gradual, it will not have a drop down to -3 dB or below before the Nyquist frequency.  The $\alpha$ at which this happens is given by the positive solution to
$$ \alpha^2+4\alpha-4 =0$$
which is
$$\alpha = -2 + 2\sqrt{2} \approx 0.82843$$
EWMA filters with an $\alpha$ that is this high, have a pole so close to the origin of the z-plane, that the filter response is so gradual, there is no frequency attenuated by 3 dB or more.
The Wikipedia derivation of the 3 dB cutoff point is flawed and comes to an incorrect/inexact result.  Specifically it directly applies a continuous time, Laplace s-domain relationship:
$$f_c = \dfrac{1}{2\pi RC}  \quad \text{(Don't use this!)}$$
into a derivation of a result for a discrete time, z-domain, filter.  That step invalidates the derivation of an exact result in the Wikipedia article.
As a consequence, the result of the derivation in the Wikipedia article is only an approximation and (subjectively) only good for cutoff frequencies up to about $0.1 F_s$.