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:


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 $\mathcal{E}$.

