过抛物线 $y^2=2px(p>0)$ 外一点 $P$ 向抛物线作两条切线，切点为 $M,N$，$F$ 为抛物线的焦点．
【难度】
【出处】
2016年全国高中数学联赛湖北省预赛
【标注】
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证明：$|PF|^2=|MF|\cdot|NF|$；标注答案略解析设 $P(x_0,y_0),M(x_1,y_1),N(x_2,y_2)$，易求得切线 $PM$ 的方程为 $y_1y=p(x+x_1)$，切线 $PN$ 的方程为 $y_2y=p(x+x_2)$．因为点 $P$ 在两条切线上，所以有$$y_1y_0=p(x_0+x_1), y_2y_0=p(x_0+x_2),$$故点 $M,N$ 均在直线 $y_0y=p(x+x_0)$ 上，所以直线 $MN$ 的方程为 $y_0y=p(x+x_0)$．
联立 $\begin{cases}y_0y=p(x+x_0),\\y^2=2px,\end{cases}$ 得 $[p(x+x_0)]^2=2pxy_0^2$，即\[x^2+2\left(x_0-\dfrac{y_0^2}p\right)x+x_0^2=0.\]由韦达定理可知：\[x_1+x_2=2\left(\dfrac{y_0^2}p-x_0\right),x_1x_2=x_0^2.\]因为 $F\left(\dfrac p 2, 0\right)$，由抛物线的第二定义可得\[|MF|=x_1+\dfrac p 2,|NF|=x_2+\dfrac p 2,\]所以\[\begin{split}|MF|\cdot |NF|&=\left(x_1+\dfrac p 2\right)\left(x_2+\dfrac p 2\right)\\&=x_1x_2+\dfrac p 2(x_1+x_2)+\dfrac{p^2}4\\&=x_0^2+\dfrac p 2 \cdot\left[2\left(\dfrac{y_0^2}p-x_0\right)\right]+\dfrac{p^2}4\\&=x_0^2+y_0^2-px_0+\dfrac{p^2}4\\&=\left(x_0-\dfrac p 2\right)^2+y_0^2\\&=|PF|^2,\end{split}\]因此 $|PF|^2=|MF|\cdot|NF|$． -
证明：$\angle PMF = \angle FPN$．标注答案略解析
方法一 因为 $\overrightarrow{FP}=\left(x_0-\dfrac p 2,y_0\right)$，$\overrightarrow{FM}=\left(x_1-\dfrac p 2,y_1\right)$，$\overrightarrow{FN}=\left(x_2-\dfrac p 2,y_2\right)$，所以\[\begin{split}\overrightarrow{FP}\cdot \overrightarrow{FM}&=\left(x_0-\dfrac p 2,y_0\right)\cdot \left(x_1-\dfrac p 2,y_1\right)\\&=x_0x_1-\dfrac p 2(x_0+x_1)+\dfrac {p^2} 4+y_0y_1\\&=x_0x_1-\dfrac p 2(x_0+x_1)+\dfrac {p^2} 4+p(x_0+x_1)\\&=x_0x_1+\dfrac p 2(x_0+x_1)+\dfrac {p^2} 4\\&=\left(x_0+\dfrac p 2\right)\left(x_1+\dfrac p 2\right),\end{split}\]又\[|MF|=x_1+\dfrac p 2,\overrightarrow{FP}\cdot \overrightarrow{FM}=|FP|\cdot|MF|\cos\angle PFM,\]所以\[\cos\angle PFM=\dfrac{\overrightarrow{FP}\cdot \overrightarrow{FM}}{|FP|\cdot|MF|}=\dfrac{\left(x_0+\dfrac p 2\right)\left(x_1+\dfrac p 2\right)}{|FP|\cdot\left(x_1+\dfrac p 2\right)}=\dfrac{x_0+\dfrac p 2}{|FP|}.\]同理可得 $\cos\angle PFN=\dfrac{x_0+\dfrac p 2}{|FP|}$．
所以 $\cos \angle PFM=\cos\angle PFN$，所以 $\angle PFM=\angle PFN$．
结合 $|PF|^2=|MF|\cdot|NF|$ 可得 $\triangle MFP \backsim \triangle PFN$，所以 $\angle PMF=\angle FPN$．方法二 只需要证明直线 $PF$ 平分 $\angle MFN$，即\[\dfrac{2\cdot \dfrac{y_0}{x_0-\dfrac p2}}{1-\left(\dfrac{y_0}{x_0-\dfrac p2}\right)^2}=\dfrac{\dfrac{y_1}{x_1-\dfrac p2}+\dfrac{y_2}{x_2-\dfrac p2}}{1-\dfrac{y_1}{x_1-\dfrac p2}\cdot \dfrac{y_2}{x_2-\dfrac p2}},\]也即\[\dfrac{2\left(x_0-\dfrac p2\right)y_0}{\left(x_0-\dfrac p2\right)^2-y_0}=\dfrac{\left(y_1y_2-p^2\right)\left(y_1+y_2\right)}{2p\left[x_1x_2-\dfrac p2\left(x_1+x_2\right)+\dfrac{p^2}4\right]-2py_1y_2},\]而\[x_1x_2=x_0^2,y_1y_2=2px_0,y_1+y_2=2y_0,\]代入即得．
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