《2025年金版教程高考科学复习解决方案数学》

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4.(2023·甘肃张掖高台县第一中学统考期末)已知点$A(x_{0}, - 2)$在抛物线$C:y^{2}=2px(p ＞ 0)$上，且$A$到$C$的焦点$F$的距离与到$x$轴的距离之差为$\frac{1}{2}$.
(1)求抛物线$C$的方程;
(2)当$p ＜ 2$时，$M$，$N$是$C$上不同于点$A$的两个动点，且直线$AM$，$AN$的斜率之积为 - 2，$AD\perp MN$，$D$为垂足.证明：存在定点$E$，使得$|DE|$为定值.
提示:同步《课时作业(A本)》P368

答案： 解　
(1)抛物线C：$y^{2}=2px(p\gt0)$的焦点为$F(\frac{p}{2},0)$，准线方程为$x=-\frac{p}{2}$，又点$A(x_{0},-2)$在抛物线C：$y^{2}=2px(p\gt0)$上，即$(-2)^{2}=2px_{0}$，$\therefore x_{0}=\frac{2}{p}$，即$A(\frac{2}{p},-2)$，依题意，可得$\frac{2}{p}+\frac{p}{2}-2=\frac{1}{2}$，解得p = 1或p = 4，$\therefore y^{2}=2x$或$y^{2}=8x$．
(2)证明：$\because p\lt2$，$\therefore y^{2}=2x$，$A(2,-2)$．设$MN:x = my + n$，$M(\frac{y_{1}^{2}}{2},y_{1})$，$N(\frac{y_{2}^{2}}{2},y_{2})$，联立$\begin{cases}y^{2}=2x\\x = my + n\end{cases}$，消去x，整理得$y^{2}-2my - 2n = 0$，$\Delta=4m^{2}+8n\gt0$，（ⅰ）且$y_{1}+y_{2}=2m$，$y_{1}y_{2}=-2n$，$\therefore k_{AM}\cdot k_{AN}=\frac{2}{y_{1}-2}\cdot\frac{2}{y_{2}-2}=-2$，$\therefore(y_{1}-2)(y_{2}-2)=-2$，即$y_{1}y_{2}-2(y_{1}+y_{2})+6 = 0$，$\therefore n + 2m = 3$，适合(ⅰ)，将$n = 3 - 2m$代入$x = my + n$，得$x - 3 = m(y - 2)$，令$\begin{cases}x - 3 = 0\\y - 2 = 0\end{cases}$，解得$\begin{cases}x = 3\\y = 2\end{cases}$，$\therefore$直线MN恒过定点$Q(3,2)$．又$AD\perp MN$，$\therefore$点D在以AQ为直径的圆上，$\because A$，Q的中点为$(\frac{5}{2},0)$，$|AQ|=\sqrt{(2 - 3)^{2}+(-2 - 2)^{2}}=\sqrt{17}$，$\therefore$以AQ为直径的圆的方程为$(x-\frac{5}{2})^{2}+y^{2}=\frac{17}{4}$，$\therefore$存在点$E(\frac{5}{2},0)$，使得$|DE|=\frac{\sqrt{17}}{2}$，为定值．