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?
 A: Yes the equation of the ellipse you have come up with is correct. One of the axes of the ellipse is $y = - x$. Now if you want to express the ellipse in the form $ ~\displaystyle \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1$, you will have to use rotation of coordinate axes. But to just find the length of latus rectum,
Note that the distance $d$ between a directrix and the near focus is $a (\frac 1 e - e)$.
The distance from $(1, -1)~$ to $~x - y + 2 = 0$ is $2 \sqrt2$. As $e = \frac 1 {\sqrt2}$,
$a = 4, b = 2 \sqrt2~$ and so length of latus rectum is $4$.
A: 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:
$\frac{X^2}8+\frac{(Y-3\sqrt 2)^2}{16}=1$
A: Once we recognize that the major axis of the ellipse is along the line $ \ y \ = \ -x \ \ , $ this can be inserted into the ellipse equation to "reduce" it to $ \ 4x^2 - 24x + 4 \ = \ 0 \ \ , $ the solutions of which are the endpoints of the major axis; the point midway between those is naturally the center of the ellipse.  (This is what Lexi Belle Fan  is describing.)
The minor axis lies on the perpendicular line through the center; with those intersections with the ellipse found from $ \ 8x^2 - 48x + 40 \ = \ 0 \ \ , $ we have the endpoints of the minor axis and can determine its length.  From the (semi-)major and (semi-)minor axis lengths, we can find the eccentricity and the focal distance $ \ c \ = \ ae \ \ , $ which then locates the foci on $ \ y \ = \ -x \ $ .
A line through either focus parallel to $ \ y \ = \ x \ $ gives us the end-points of the latus rectum , from which we can find their separation distance; the associated quadratic equation is $ \ 8x^2 -16x - 8 \ = \ 0 \ \ . $
Since all the lengths are hypotenuses of right isosceles triangles, even calculating the values by hand is reasonably simple. I am listing the quadratic equations as check-points along the way, and leaving you to fill in the details.  This problem (fortunately -- or maybe intentionally) has a "nice enough" rotation of axes that it doesn't really require the additional processing to answer the question.
ADDENDUM -- Since the problem gives you one focus of the ellipse, you can pretty much "skip down" to finding the intersections of the line parallel to $ \ y \ = \ x \ $ (perpendicular to the major axis) passing through $ \ (1 \ , \ -1) \ $ and computing the distance between them.  You don't require any other measures of the ellipse, unless there are other questions about it.
To answer a possible question about how we know the major axis is on $ \ y \ = \ -x \ \ , $ the coefficients in your equation  $ \ 3x^2 + 3y^2 + 2xy - 12x + 12y + 4 \ = \ 0 \ \ $ are a "tip-off".  We can show that this ellipse has "diagonal symmetry":  if a point $ \ (x \ , \ y) \ $ is on its graph, so is the point $ \ (-x \ , \ -y) \ \ . $
