952

已知抛物线 ${\displaystyle C}$ 的方程为 ${\displaystyle y={\displaystyle \frac{1}{4}{x}^2}}$, ${\displaystyle F}$ 为其焦点，点 ${\displaystyle N}$ 坐标为 ${\displaystyle (0,-4)}$, 过点 ${\displaystyle F}$ 作直线交抛物线 ${\displaystyle C}$ 于 ${\displaystyle A,B}$ 两点，${\displaystyle D}$ 是 ${\displaystyle x}$ 轴上一点，且满足 ${\displaystyle |DA|=|DB|=|DN|}$, 则直线 ${\displaystyle AB}$ 的斜率为 ${\displaystyle \mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}}$.

设 ${\displaystyle l_{AB}:y=kx+1}$; 由于 ${\displaystyle D}$ 是 ${\displaystyle \mathrm{\Delta }ABN}$ 的外心，且过 ${\displaystyle N(0,-4)}$, 故圆 ${\displaystyle D:(x-t{)}^2+y^2=t^2+16}$. 这样 ${\displaystyle C}$ 与圆 ${\displaystyle D}$ 的交点曲线系为$$\begin{array}{r}(x^2-2tx+y^2-16)+\lambda (y-\frac{1}{4}{x}^2)=0,\end{array}$$
${\displaystyle y-kx-1=0}$是它的一个因式，进而消去${\displaystyle y}$得$$\begin{array}{r}(k^2+1-\frac{\lambda }{4})x^2+(2k-2t+\lambda k)x+\lambda -15=0,\end{array}$$
令各项系数为${\displaystyle 0}$, 解得${\displaystyle k={\displaystyle \pm \frac{\sqrt{11}}{2}}}$.