计算有技巧，却难在因式分解
已知椭圆\(C:\dfrac{x^2}{8}+\dfrac{y^2}{4}=1\)，过点\((1,0)\)的直线与\(C\)相交于\(A,B\)两点，过点\(C\)上的点\(P\)作\(x\)轴的平行线交线段\(AB\)于点\(Q\)，直线\(OP\)的斜率为\(k^{\prime},\triangle APQ\)的面积为\(S_1,\triangle BPQ\)的面积为\(S_2\)，若\(|AP|\cdot S_2=|BP|\cdot S_1\)，设\(l\)的斜率为\(k\)，判断\(k\cdot k^{\prime}\)是否为定值，并说明理由.
解.
面积关系转化为:\(|AP|\cdot |QP|\cdot |BP|\sin\angle BPQ=|BP\cdot|QP|\cdot|AP|\cdot\sin\angle APQ\)
得\(\angle BPQ=\angle APQ\) ，即\(k_{BP}+k_{AP}=0\)
设\(A(x_1,y_1),B(x_2,y_2),P(x_0,y_0)\)
联立直线\(my+1=x\)与椭圆得\(y^2(2+m^2)+2my-7=0\)
有\(y_1+y_2=-\dfrac{2m}{2+m^2},y_1y_2=\dfrac{-7}{2+m^2}\)
\(k_{BP}+k_{AP}=\dfrac{y_2-y_0}{x_2-x_0}+\dfrac{y_1-y_0}{x_1-x_0}=\dfrac{(y_2-y_0)(x_1-x_0)+(y_1-y_0)(x_2-x_0)}{(x_2-x_0)(x_1-x_0)}=0\)
联立\(my+1=x\)整理得
\(2my_1y_2+(y_1+y_2)(1-x_0-my_0)-2y_0(1-x_0)=0\)
带入\(y_1+y_2,y_1y_2\)得
\(2m\cdot \dfrac{-7}{2+m^2}-\dfrac{2m}{2+m^2}(1-x_0-my_0)-2y_0(1-x_0)=0\)
整理有:\(m(x_)-8)+m^2y_0-y_0(2+m^2-2x_0-m^2x_0)=0\)
整理有:\(m(x_0-8)+y_0(m^2x_0+2x_0-2)=0\)
整理有:\(m(x_0-8)+2y_0(x_0-1)+m^2y_0x_0=0\)
又因\(P\)在椭圆上，满足\(\dfrac{x_0^2}{8}+\dfrac{y_0^2}{4}=1\)
即\(8=x_0^2+2y_0^2\)，代回有
\(m(x_0-x_0^2-2y_0^2)+2y_0(x_0-1)+m^2y_0x_0=0\)
分解因式得
\((2y_0-mx_0)(x_0-my_0-1)=0\)
因\(AB\)不过\(P\)点，所以\(x_0-my_0-1\neq 0\)
从而\(2y_0-mx_0=0\)，即\(\dfrac{y_0}{x_0}=\dfrac{m}{2}\)
从而\(k\cdot k^{\prime}=\dfrac{m}{2}\cdot \dfrac{1}{m}=\dfrac{1}{2}\)