隐藏的斜率和问题
已知双曲线\(C\)为\(\dfrac{x^2}{2}-y^2=1\)，直线\(l\)交\(C\)于\(P,Q\)两点.若直线\(AP,AQ\)与\(y\)轴分别相交于\(M,N\)两点，且\(\overrightarrow{OM}+\overrightarrow{ON}=\overrightarrow{0}\)，证明：\(l\)过定点.
解
设\(M(m,0)\)则由题\(M(0,-m)\)
从而有\(k_{AP}=\dfrac{1-m}{2},k_{AG}=\dfrac{1+m}{2}\)
即\(k_{AP}+k_{AG}=1\)
则此题是一个常考模型，只是把斜率和、积为定值的这个前置条件隐藏了
设\(AP:y-1-k_{AP}(x-2)=0,AG:y-1-k_{AG}(x-2)=0\)
两边相乘有
\[(y-1)^2-(y-1)(x-2)+k_{AP}k_{AG}(x-2)^2=0----(1)
\]
又因双曲线\(C\)可以写成\(\dfrac{(y-1)(y+1)}{(x-2)(x+2)}=\dfrac{1}{2}\)
从而\(y-1=\dfrac{(x-2)(x+2)}{2(y+1)}\)
带回\((1)\)有
\[\dfrac{(x-2)^2(x+2)^2}{4(y+1)^2}-\dfrac{(x-2)(x+2)}{2(y+1)}(x-2)+k_{AP}k_{AG}(x-2)^2=0
\]
即
\[(x+2)^2-2(x+2)(y+1)+4k_{AP}k_{AG}(y+1)^2=0----(2)
\]
则\(-2\cdot (1)+(2)\)消去\(xy\)有
\[(x^2-2y^2+4x+4y+2)+2(-4y-2x)+2k_{AP}k_{AQ}(2y^2-x^2+4y+4x-2)=0
\]
联立\(C:\dfrac{x^2}{2}-y^2=1\)即\(x^2-2y^2=2\)有
\[(4x+4y+2-2)+2(-4y-2x)+2k_{AP}k_{AQ}(2+4y+4x-2)=0
\]
整理得\(PQ\)方程
\[-4y+4+2k_{AP}k_{AQ}(4y+4x)=0
\]
从而过定点\((0,1)\)