The one end of the latusrectum of the parabola ${ y }^{ 2 }-4x-2y-3=0$ is at

Given equation is ${ y }^{ 2 }-4x-2y-3=0$

$\Rightarrow \left( { y }^{ 2 }-2y \right) =4x+3$

$\Rightarrow { \left( y-1 \right) }^{ 2 }-1=4x+3$

$\Rightarrow { \left( y-1 \right) }^{ 2 }=4\left( x+1 \right)$

Shift the origin to $\left( -1,1 \right)$

$\Rightarrow { Y }^{ 2 }=4X$

Here focus is at $\left( 1,0 \right)$.

Hence, focus of original parabola becomes $\left( 1-1,0+1 \right) =\left( 0,1 \right)$

Therefore, equation of latusrectum is $x=0$.

Thus point of intersection of parabola and latus rectum is

${ y }^{ 2 }-2y-3=0\Rightarrow y=-1$ or $3$

So, the required points are $\left( 0,-1 \right) ,\left( 0,3 \right)$.