The ends of latusrectum of a parabola are (–3, 1) and (1, 1) then equation of parabola is

Given the ends of the latursrectum are $L\left(-3,1\right),L\text{'}\left(1,1\right)$

Since $y-$ coordinates of the latusrecturm are equal so axis is parallel to y-axis then the equation of the parabola can be taken as ${\left(x-h\right)}^2=4a\left(y-k\right)$

The length of the latusrectum is 4 units, it implies that $4a=4$

The focus is midpoint of $LL\text{'}$

Hence, $S\left(-1,1\right)$

Hence the equation of the required parabola is ${\left(x+1\right)}^2=4\left(y-k\right)$

It is passing through the point $\left(1,1\right)$

$4=4\left(1-k\right)\Rightarrow k=0$

Therefore, the equation of the parabola is $\boxed{(\mathrm{x}+1{)}^2=4\mathrm{y}}$