过抛物线 $x^{2}=4 y$ 的焦点 $F$ 作直线, 交抛物线于 $P_{1}\left(x_{1}, y_{1}\right), P\left(x_{2}, y_{2}\right)$ 两点, 若 $y_{1}+y_{2}=6$,则 $\left|P_{1} P_{2}\right|=(\quad)$
$x^{2}=4 y$ 的焦点为 $(0,1)$, 设过焦点 $(0,1)$ 的直线为 $y=k x+1$, 则令 $k x+1=rac{x^{2}}{4}$, 即 $x^{2}-4 k x-4=0$,由韦达定理得 $\mathrm{x}_{1}+\mathrm{x}_{2}=4 \mathrm{k}, \mathrm{x}_{1} \mathrm{x}_{2}=-4, \mathrm{y}_{1}=\mathrm{kx}_{1}+1, \mathrm{y}_{2}=\mathrm{kx}_{2}+1$, 所以 $\mathrm{y}_{1}+\mathrm{y}_{2}=\mathrm{k}\left(\mathrm{x}_{1}+\mathrm{x}_{2}
ight)+2=4 \mathrm{k}^{2}+2=6$, 所以 $\mathrm{k}^{2}=1$,所以 $|A B|=\left|x_{1}-x_{2}
ight| \sqrt{k^{2}+1}=\sqrt{\left(k^{2}+1
ight)\left[\left(x_{1}+x_{2}
ight)^{2}-4 x_{1} x_{2}
ight]}=\sqrt{2\left(16 k^{2}+16
ight)}=\sqrt{2 	imes 32}=8$.