One of the asymptotes of hyperbola is $2x-y =5$ and one focus is $(2,4)$. It's eccentricity is 2. Find its latus rectum.
My attempt
We know that the angle between an asymptote and the line joining focus and centre is $$\arccos \frac1e =\arccos \frac12 = \frac\pi3 $$.
To rotate the given asymptote by an angle of $\frac\pi3$, I replaced $x$ and $y$ with $x'$and $y'$ where $$\begin{bmatrix}x'\\y' \end{bmatrix}=\begin{bmatrix}\frac12 & -\frac{\sqrt3}{2} \\ \frac{\sqrt3}{2} & \frac12 \end{bmatrix}\begin{bmatrix}x\\y \end{bmatrix}$$ which yields $$\begin{bmatrix}x'\\y' \end{bmatrix}=\begin{bmatrix}\frac x2 -\frac{\sqrt3}{2}y\\\frac{\sqrt3}{2}x+\frac y2 \end{bmatrix}$$
Substituting this into the equation for asymptote, we get the rotated line as:
$$x(1-\frac{\sqrt3}{2}) -y(\sqrt3 + \frac12) + k =0$$
This line should pass through the given focus $(2,4)$, substituting these coordinates we obtain $k=5\sqrt3$, so the equation of line joining focus and centre is $$x(1-\frac{\sqrt3}{2}) -y(\sqrt3 + \frac12) + 5\sqrt3=0$$
I got stuck here, and don't really know how to proceed. Can somebody help solve it further?
PS: I'm still in highschool, so it would be appreciated if you do not use highly advanced concepts to solve this.
Thank you.