经典\(e^2-1\)应用
已知椭圆\(M:\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1(a>b>0)\)的左右顶点为\(A\)、\(B\)，\(P\)是椭圆上异于\(A\)、\(B\)的动点，满足\(k_{PB}\cdot k_{PB}=-\dfrac{1}{4}\)，当\(P\)是当顶点时，\(\triangle ABP\)面积为\(2\)
\((1)\) 求\(M\)方程
\((2)\) 若直线\(AP\)交直线\(l:x=4\)于点\(C\)，直线\(CB\)交椭圆于\(Q\)点，证明：\(PQ\)过定点.
解
\((1)\) \(\begin{cases} -\dfrac{b^2}{a^2}=-\dfrac{1}{4}\\ \\ ab=2 \end{cases}\Rightarrow a=2,b=1\).从而\(C:\dfrac{x^2 }{4}+y^2=1\)
\((2)\) 由椭圆第三定义得\(k_{AP}\cdot k_{PB}=-\dfrac{1}{4}\)
设\(C:(4,m)\)，则\(k_{AP}=\dfrac{m}{6},k_{BQ}=k_{CQ}=\dfrac{m}{2}\)
从而\(k_{AP}=\dfrac{1}{3}k_{BQ}\)
即\(\dfrac{1}{3}k_{BP}\cdot k_{BQ}=-\dfrac{1}{4}\)
即\(k_{BP}\cdot k_{BQ}=-\dfrac{3}{4}\)
设\(BQ:y=k_{BQ}(x-2),BP:y=k_2(x-2)\)
即\(BQ:y-k_{BP}(x-2)=0,BP:y-k_2(x-2)=0\)
两直线相乘有
\(y^2-y(x-2)(k_{BQ}+k_{BP})+k_{BQ}k_{BP}(x-2)^2=0\)
因\(k_{BP}\cdot k_{BQ}=-\dfrac{3}{4}\)
从而\(y^2-y(x-2)(k_{BQ}+k_{BP})-\dfrac{3}{4}(x-2)^2=0\)
联立\(\dfrac{x^2 }{4}+y^2=1\)有\(y^2=\dfrac{(2-x)(2+x)}{4}\)
带回有
\(\dfrac{(2-x)(2+x)}{4}-y(x-2)(k_{BQ}+k_{BP})-\dfrac{3}{4}(x-2)^2=0\)
即得到\(PQ\)直线： \(-\dfrac{2+x}{4}-y(k_{BQ}+k_{BP})-\dfrac{3}{4}(x-2)=0\)
即\(-x+1-y(k_{BQ}+k_{BP})=0\)
过定点\((1,0)\).