How to find the length of the latera recta of the ellipse $3x^2+3y^2+2xy-12x+12y+4=0$? Context:
"Find the equation of the ellipse one of whose focus is $(1,-1)$ and the equation of whose directrix is $x-y+2=0$ and eccentricity is $\frac{1}{\sqrt{2}}$. Find also the length of its latera recta."
I was able to find the equation of the ellipse correctly, but I'm unable to find the length of the latera recta.

$$3x^2+3y^2+2xy-12x+12y+4=0$$
The given ellipse can't be expressed in the form $\frac{(x-\alpha)^2}{a^2}+\frac{(y-\beta)^2}{b^2}=1$. If that was possible, I could've found out the length easily using $\frac{2a^2}{b}$ or $\frac{2b^2}{a}$.
My book only taught me how to find the length of the latera recta when the ellipse can be expressed in the form $\frac{(x-\alpha)^2}{a^2}+\frac{(y-\beta)^2}{b^2}=1$, so I have no idea why it's asking me to find the length of the latera recta in this case.
What do I do now?

The equation can be re-written as
$\begin{bmatrix}x&y\end{bmatrix}\color{blue}{\begin{bmatrix}3&1\\1&3\end{bmatrix}}\begin{bmatrix}x\\y\end{bmatrix}+\begin{bmatrix}-12&12\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}+4=0\tag 1$
Now, by diagonalization of blue colored matrix:
$\overbrace{\begin{bmatrix}3&1\\1&3\end{bmatrix}}^{M}=\overbrace{\begin{bmatrix}\frac 1{\sqrt 2}& \frac 1{\sqrt 2}\\\frac 1{\sqrt 2}&-\frac 1{\sqrt 2}\end{bmatrix}}^{P}\overbrace{\begin{bmatrix}4&0\\0&2\end{bmatrix}}^{D}\begin{bmatrix}\frac 1{\sqrt 2}& \frac 1{\sqrt 2}\\\frac 1{\sqrt 2}&-\frac 1{\sqrt 2}\end{bmatrix}$
So $M=PDP^{-1}$. Substitute for M in $(1)$ and put $\begin{bmatrix}X\\Y\end{bmatrix}:=\begin{bmatrix}\frac 1{\sqrt 2}& \frac 1{\sqrt 2}\\\frac 1{\sqrt 2}&-\frac 1{\sqrt 2}\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}$ to transform $(1)$ to the following familiar form of the ellipse:
$\frac{X^2}8+\frac{(Y-3\sqrt 2)^2}{16}=1$
