We have the next equation
[tex]y=x^2+6[/tex]First, we need to find the focus of this parabola the vertice is in (0,6)
[tex]4p\mleft(y-k\mright)=\mleft(x-h\mright)^2[/tex]where in our case h =0, k=6
[tex]4\cdot\frac{1}{4}(y-6)=x^2[/tex]Therefore the focus will be
[tex](0,6+\frac{1}{4})=(0,\frac{25}{4})[/tex]Then for Latus Rectum is located between the next points
[tex](-0.5,\frac{25}{4})\text{ and (}0.5,\frac{24}{5}\text{)}[/tex]the latus Rectum
[tex]4p=4(\frac{1}{4})=1[/tex]the length of the latus rectum is 1
