955

已知抛物线 ${\displaystyle C:y^2=2x}$ 的焦点为 ${\displaystyle F}$, ${\displaystyle M,N}$ 为 ${\displaystyle C}$ 上满足 ${\displaystyle \overrightarrow{MF}\cdot \overrightarrow{NF}=0}$ 的两点，求 ${\displaystyle |MF|+|NF|}$ 的最小值.

${\displaystyle F{\displaystyle (\frac{1}{2},0)}}$. 设 ${\displaystyle M(2m^2,2m),N(2n^2,2n)}$, 则由条件得$$\begin{array}{r}(2m^2-\frac{1}{2})(2n^2-\frac{1}{2})+2m\cdot 2n=0\Rightarrow {(2mn+\frac{1}{2})}^2=(m-n{)}^2.\end{array}$$
不妨设${\displaystyle {\displaystyle 2mn+\frac{1}{2}=m-n}}$, 则$$\begin{array}{rl}& m-n=2mn+\frac{1}{2}\ge -\frac{1}{2}(m-n{)}^2+\frac{1}{2}\\ \Rightarrow & m-n\in (-\mathrm{\infty },-\sqrt{2}-1]\cup [\sqrt{2}-1,+\mathrm{\infty }),\end{array}$$
从而$$\begin{array}{rl}MF+NF& =2m^2+\frac{1}{2}+2n^2+\frac{1}{2}=(m+n{)}^2+(m-n{)}^2+1\\ & \ge 0+(\sqrt{2}-1{)}^2+1=4-2\sqrt{2}.\end{array}$$
可以有取等条件${\displaystyle {\displaystyle (m,n)=(\frac{\sqrt{2}-1}{2},-\frac{\sqrt{2}-1}{2})}}$.