If you are familiar with eigenvalues and eigenvectors, and some linear-algebra, following should be pretty much straight forward. Otherwise, you should look for other answers. Implementing them in code should also be easy in any language. I leave some of the details to be filled in.
The basic idea is that every ellipse is a rotated, axis-scaled and translated version of a circle. Axis-scaled in the sense, each axis will be given a different scaling. So once you find the endpoints of this circle, you work backwards, through all this operations (in the reverse order) to find the endpoints of the ellipse.
You can rewrite your ellipse equation as
\begin{align}
x^{T}Qx-r^{T}x+f=0
\end{align}
where $Q$ is the symmetric $2 \times 2$ matrix
\begin{align}
Q=\begin{bmatrix}A & \frac{B}{2} \\ \frac{B}{2} & C\end{bmatrix}
\end{align}
and $r$ is the $2 \times 1$ vector
\begin{align}
r=\begin{bmatrix}-D \\ -E\end{bmatrix}
\end{align}
Consider eigen decomposition of $Q=U^{T}DU$ where $D$ is diagonal matrix and contains its eigenvalues $\lambda_1,\lambda_2$, and $U$ is the matrix containing eigenvectors. Define $y=Ux$. Then your ellipse can be written as
\begin{align}
y^{T}Dy-v^{T}y+f=0
\end{align}
where $v=Ur$. Define $z=D^{\frac{1}{2}}y$ where $D^{\frac{1}{2}}$ is a diagonal matrix with diagonal entries as $\sqrt{\lambda_1},\sqrt{\lambda_2}$. Thus, your ellipse equation should become
\begin{align}
z^{T}z-w^{T}z+f=0
\end{align}
where $w=D^{\frac{-1}{2}}v$.
If $z=[z_1,z_2]^{T}$, the last equation can be written as
\begin{align}
(z_1-X_c)^{2} + (z_2-Y_c)^{2}=a^{2}
\end{align}
This you can do, using completion of squares for each terms $z_1$ and $z_2$.
This is a circle with radius $a$ and center $(X_c,Y_c)$. The endpoints of the major axis are $(a,0)$ and $(-a,0)$. Now, you have to work backwards. First you shift the endpoints by adding $(X_c,Y_c)$. Then, you scale back using $D^{\frac{1}{2}}$ and then rotate using $U^{T}$.
You need to center the conic to get the correct RHS constant.
From $\frac349.24^2\approx\frac547.16\approx64$, I conclude that it must be $16$.
Best Answer
The length of a semi major axis is just $b$, if the equation of the ellipse is $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ where $a<b$ which here is $\frac5{\sqrt3}$