If the locus of the point $(4t^2 - 1, 8t-2)$ represents a parabola then the equation of latus rectum is

$x=4t^2-1,y=8t-2$

$x+1=4t^2,y+2=8t$

let $X=x+1;Y=y+2$

then the new coordinates will be $X=4t^2, Y=8t$

$Y^2=4aX$ is the general equation of a parabola

$(8t)^2=4a\times 4t^2$

$64t^2=16at^2\Rightarrow a=4$

Equation of latus rectum of $Y^2=4aX$ is $X=a$

But we know that $X=x+1$ ,substituting that in $X=a$

$x+1=4\Rightarrow x-3=0$