Equilateral triangle inscribed in an ellipse An equilateral triangle of side length $ L\approx 6.14$ and one side inclination $ \approx 49.52^{\circ}$ is inscribed in an ellipse of semi-axes $(a,b) = (5,3)$. Drawn in Geogebra by Java mousing .. trial/error.
 
Are  $ (L,\alpha) = f(a,b) $ in this configuration unique? If so, what is the exact length and a side slope as a function of $(a,b)?$. If not, what are all equilateral triangles in a set that can be drawn on  parameter $\alpha$ or any other convenient parameter? 
Thanking you in advance for helpful suggestions.
EDIT1:
Starting with a smaller ellipse $ (a,b)=(1.5,1) $ and a single point on it as equilateral triangle center locus $(u,v)=(1,0.74)$ on $(a,b)=(13.43,10.95)$ of a larger ellipse results in radius $11.94$ according to answer by achille hui. A single equilateral triangle (of such inverse procedure) is sketched here:

 A: (Totaly modified answer : the previous one had no longer any interest).
I would like to address here the problem in a kind of dual equivalent form (see Fig. 1) :
Being given a fixed equilateral triangle $T$ (I will consider here vertices $A_0(2,0),A_1(-1,\sqrt{3}),A_2(-1,-\sqrt{3})$) how can be described the family of ellipses that are circumscribed to $T$ and have a given eccentricity (or, equivalently have a fixed ratio $B/A$, here $3/5$, of the half-lengthes of their axes) ?
It can be solved in a rather easy way.


*

*Firstly, the general equation of such a conic curve can be written :


$$\tag{1}ax^2+2bxy+cy^2+2dx+2ey-4=0$$
(the constant term is necessarily non zero, otherwise the conic curve would be passing through the origin $O$ : but it is impossible for an ellipse, a convex curve, to pass through points $A_0,A_1,A_2$ and $O$ where $O$ is strictly inside triangle $T$; the non zero constant has been chosen such in order to ease further calculations).
Being constrained to pass through the points $A_k$, (1) can be reduced to $5-3=2$ free parameters. One can easily check that, keeping parameters $c$ and $b$, the resulting form is :
$$\tag{2}(2-c)x^2+2bxy+cy^2+2(c-1)x+2by-4=0$$
with associated matrix
$$\tag{3}M=\begin{pmatrix}(2-c)&b&(c-1)\\ b&c&b\\ (c-1)&b&-4\end{pmatrix}$$


*

*Now, how can be expressed the constraint $B/A=3/5$ ?


It is well known that this ratio is equal to $\sqrt{\lambda_1/\lambda_2}$ where $0<\lambda_1<\lambda_2$ are the eigenvalues of the $2 \times 2$ upper left block of $M$ given in (3) (eigenvalues positivity is due to the fact that matrix $M$ is a positive-defined matrix).
The characteristic equation of this block is $\begin{vmatrix}2-c-\lambda&b\\b&c-\lambda\end{vmatrix}=0$ ; its roots are :
$$\lambda_k=1\pm\sqrt{E} \ \ \ \text{with} \ \ \ \ E:=1+b^2+c(c-2).$$
Condition
$$\dfrac{\lambda_1}{\lambda_2}=\dfrac{1-\sqrt{E}}{1+\sqrt{E}}=\left(\dfrac35\right)^2$$
gives, after elementary calculations, the following expression for $b$ in terms of $c$ :
$$b=\pm\sqrt{c(2-c)-225/289}=\pm\sqrt{-(c-9/17)(c-25/17)}$$
(under condition $9/17\leq c \leq 25/17$).
As a conclusion, going back to the original problem, one can say that there is an infinite number of equilateral triangles that can be inscribed in the given ellipse, or any other ellipse with a different ratio $A/B$ in fact because the method we have used is general. 
[Many thanks to @achille hui for his very constructive comments]
Not surprisingly, generated ellipses give a set of solutions wich is invariant by $k 2 \pi/3$ rotations.

Matlab program:
 clear all;close all;hold on;axis equal;axis off
 s1='*(-x.^2+2*x+y.^2)+';
 s2='*2*y.*(x+1)+2*x.^2-2*x-4';
 plot([2,-1,-1,2],[0,sqrt(3),-sqrt(3),0],...
     'linewidth',2,'color','k');
 f=@(n)(num2str(n));
 for k=9:25
        c=k/17;
        b=sqrt(-(c-9/17)*(c-25/17));
        for p=1:2
            ez=ezplot([f(c),s1,f(b),s2]);%implicit plot
            set(ez,'linecolor',[k==18,0,1])
            set(ez,'linewidth',(k==18)+1)
            b=-b;
        end;
 end;

A: In following discussion, we will assume $a > b > 0$.
Identify the plane with complex plane through the map $\mathbb{R}^2 \ni (x,y) \mapsto z = x + iy\in \mathbb{C}$.
In terms of $z$, the equation of ellipse $\mathcal{E} : \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ becomes
$$A(z^2+\bar{z}^2) + Bz\bar{z} = 1\quad\text{where}\quad
\begin{cases}
A = \frac{1}{4a^2} - \frac{1}{4b^2}\\
B = \frac{1}{2a^2} + \frac{1}{2b^2}
\end{cases}
$$
Let $\omega = e^{i\frac{2\pi}{3}}$ be the cubic root of unity. There is a 3-to-1 parameterization of equilateral triangles in the plane using two complex number
$p, q$. $p$ is the center and $q$ is the difference between one of the vertices and $p$. The vertices of the triangle will be located at $p + q\omega^k$ for $k = 0, 1, 2$.
In order for such a triangle to lie on ellipse $\mathcal{E}$, we need
$$A\left((p+q\omega^k)^2 + (\bar{p} + \bar{q}\omega^{-k})^2\right)
+ B(p+q\omega^k)(\bar{p} + \bar{q}\omega^{-k}) = 1 \quad\text{ for }\quad k = 0,1,2$$
Separating coefficients of different $\omega^k$, one obtain
$$\begin{align}
A (p^2 + \bar{p}^2) + B( p\bar{p} + q\bar{q}) = 1\tag{*1a}\\
A(2pq+ \bar{q}^2) + B\bar{p}q = 0\tag{*1b}\\
A(2\bar{p}\bar{q} + q^2) + Bp\bar{q} = 0\tag{*1c}
\end{align}
$$
Equation $(*1c)$ give us nothing new, it is just a complex conjugation of $(*1b)$.
By rewriting equation $(*1b)$ as $(2Ap+B\bar{p})q + A\bar{q}^2 = 0$, we find
$$|q| = \left|2p + \frac{B}{A}\bar{p}\right|\quad\text{ and }\quad
(2Ap + B\bar{p})q^3 + A|q|^2 = 0\tag{*2}
$$
The equation on the left tell us once $p$ is known, so do $|q|$. We then use equation on the right to determine $q^3$ and hence $q$ up to a power of $\omega$. In order for this compatible with $(*1a)$, the necessary and sufficient condition is
$$A(p^2 + \bar{p}^2) + B\left( p\bar{p} + \left(2p + \frac{B}{A}\bar{p}\right)\left(2\bar{p} + \frac{B}{A}p\right)\right) = 1$$
With some algebra, one can simplify above to
$$\frac{u^2}{a_1^2} + \frac{v^2}{b_1^2} = 1\quad\text{ where }\quad
\begin{cases}
p  &= u + iv\\
a_1 &= a\frac{a^2-b^2}{a^2+3b^2}\\
b_1 &= b\frac{a^2-b^2}{b^2+3a^2}
\end{cases}
$$
This is the equation for another ellipse $\mathcal{E}_1$. Pick any point $p = u+iv$ on this ellipse, the corresponding $|q|^2$ is given by the equation
$$B|q|^2 = 1 - ( A(p^2+\bar{p}^2) + Bp\bar{p}) = 1 - \left(\frac{u^2}{a^2} + \frac{v^2}{b^2}\right)$$
If one use this $|q|$ as radius and draw a circle centered at $p$, it will intersect ellipse $\mathcal{E}$ at $3$ or $4$ points ($3$ when $uv = 0$, $4$ otherwise). Three of them will
form an equilateral triangle (one can use $(*2)$ to figure out exactly which three are them).
Let $q = |q|e^{i\theta}$, we can parameterize the family of equilateral
triangles inscribed in $\mathcal{E}$ using $\theta$.
Let $\lambda(\theta)$ be following horrible expression
$$\lambda(\theta) = \frac{1}{\sqrt{\frac{a^4}{a_1^2}\cos(3\theta)^2 + \frac{b^4}{b_1^2}\sin(3\theta)^2}}$$
For any $\theta$, one can verify following three points $z_1,z_2,z_3$ 
$$z_k = \lambda(\theta)\left(a^2\cos(3\theta) + b^2\sin(3\theta)i -\frac1A e^{i(\theta + \frac{2\pi}{3}k)}\right)
\quad\text{ for }\quad k = 0,1,2$$
all lie on $\mathcal{E}$ and hence define an equilateral
equilateral triangle inscribed in $\mathcal{E}$.
An implementation of this mess in GeoGebra indicate as $\theta$ varies over $[0,2\pi)$, $z_0(\theta)$ will covers all points on $\mathcal{E}$ from one to three times. When $\theta \sim \frac{\pi}{2} \leftrightarrow
z_0(\theta) \sim b i\;$ or $\;\theta \sim \frac{3\pi}{2} \leftrightarrow z_0(\theta) \sim -b i$, $z_0(\theta)$ is moving backward!
Following animation illustrates what happens for the configuration $(a,b) = (5,3)$. $P$ is the triangle center and the unlabelled point is $z_0(\theta = {\rm th})$.

As one can see,


*

*This parameterization gives us a 3-to-1 parametrization of the family of equilateral triangles inscribed in $\mathcal{E}$.

*For points on $\mathcal{E}$ near $(0,\pm b)$, it is possible to have more than one equilateral triangles having that points as vertex!


Update
@Ng Chung Tak has pointed out the fourth intersection is located at 
$$z_4 = \lambda (\theta) \left( a^2\cos 3\theta+ib^2 \sin 3\theta-\dfrac{e^{-3i\theta}}{A} \right)$$
This result leads to a more geometric way to construct equilateral triangles inscribed in $\mathcal{E}$.



*

*Pick any point $A$ on $\mathcal{E}_1$ on first quadrant. Reflect $A$ with respect to $x$-axis to get $A'$.

*Let $B$ be the intersection of ray $OA'$ with $\mathcal{E}$.

*Construct a line through $A$ parallel to the normal line of $\mathcal{E}$ at $B$.

*Let $C$ be the intersection of this line with $\mathcal{E}$ in fourth quadrant.

*Construct a circle centered at $A$ through $C$. Let $D, E, F$ be the other intersection points with $\mathcal{E}$.

*$\triangle DEF$ will be an equilateral triangle inscribed in $\mathcal{E}$.

