$y^{2}-4 x-2 y-3=0$
$\Rightarrow \left(y^{2}-2 y\right)=4 x+3$
$\Rightarrow (y-1)^{2}-1=4 x+3$
$\Rightarrow (y-1)^{2}=4(x+1)$
Shift the origin to $(-1,1)$. $\Rightarrow \quad Y^{2}=4 X$
$\Rightarrow $ Focus is at $(1,0)$.
Hence, focus of original parabola becomes $(1-1,0+1)=(0,1)$
$\therefore $ Equation of latusrectum is $x=0$
$\therefore $ Point of intersection of parabola and latusrectum is
$y^{2}-2 y-3=0 \Rightarrow y=-1 $ or }$3$
So, required points are $(0,-1),(0,3)$.