如图，过椭圆 $a x^2 +b y^2 = 1 (b>a>0)$ 中心 $O$ 的直线 $l_{1}$、$l_{2}$ 分别交椭圆于点 $A$、$E$、$B$、$G$ 四点，且直线 $l_{1}$、$l_{2}$ 的斜率之积是 $-\dfrac{a}{b}$，过点 $A$、$B$ 作两条平行线 $l_{3}$、$l_{4}$，设 $l_{2}\cap l_{3} = M$，$l_{1}\cap l_{4} = N$，且 $CD \cap MN=P$，求证：$OP \parallel l_{3}$．
【难度】
【出处】
2015年全国高中数学联赛河南省预赛
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【答案】
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【解析】
要证 $OP \parallel l_{3}$ 等价于 $\dfrac{ON}{OA} =\dfrac{NP}{PM}$，由 $l_{3} \parallel l_{4}$，得 $\dfrac{ON}{OA} =\dfrac{NB}{MA}$，$\dfrac{NP}{PM} =\dfrac{ND}{MC}$．
故 $\dfrac{ON}{OA} =\dfrac{NP}{PM}$ 等价于 $\dfrac{NB}{MA}=\dfrac{ND}{MC} $ 等价于 $\dfrac{ND}{NB}=\dfrac{MC}{MA}.$
设 $\dfrac{MC}{MA} = \lambda _{1}$，$\dfrac{ND}{NB}=\lambda_{2}$，问题等价于证明 $\lambda _{1}=\lambda _{2}$．
设 $A(x_{1},y_{1})$、$B(x_{2},y_{2})$、$C(x_{3},y_{3})$、$D(x_{4},y_{4})$，则由 $k_{l_{1}} \cdot k_{l_{2}} = -\dfrac{a}{b}$，得 $ax_{1}x_{2} + by_{1}y_{2}=0$．由于 $\dfrac{OM}{OB}=\dfrac{OA}{ON} $，故可设 $M(tx_{2},ty_{2})$、$N(\dfrac{x_{1}}{t},\dfrac{y_{1}}{t})$，由 $\dfrac{MC}{MA} = \lambda _{1}$，可得 $\overrightarrow {MC} = \lambda _{1}\overrightarrow {MA}.$
即 $(x_{3} - tx_{2},y_{3} - ty_{2}) = \lambda _{1}(x_{1} - tx_{2},y_{1} - ty_{2})$，则$$x_{3} = \lambda _{1}x_{1} + (1-\lambda _{1})tx_{2},y_{3} =\lambda _{1}y_{1} + (1-\lambda _{1})ty_{2}.$$将 $C(x_{3},y_{3})$ 代入椭圆：$ax^2 +by^2 = 1$ 方程可得$$\lambda_{1}^2(ax_{1}^2 + by_{1}^2) + 2\lambda_{1}(1-\lambda_{1})t(ax_{1}x_{2} + by_{1}y_{2}) + (1-\lambda_{1})^2 t^2(ax_{2}^2 +by_{2}^2)=1,$$即 $\lambda_{1}^2 + (1-\lambda_{1})^2t^2 = 1$，得 $\lambda_{1} = \dfrac{t^2 - 1}{t^2 + 1} .$
同理由 $\dfrac{ND}{NB}=\lambda_{2}$，可得 $\overrightarrow {ND} = -\lambda_{2}\overrightarrow {NB}$，即$$(x_{4} - \dfrac{x_{1}}{t},y_{4} - \dfrac{y_{1}}{t})=-\lambda_{2}(x_{2} - \dfrac{x_{1}}{t},y_{2} - \dfrac{y_{1}}{t}).$$整理可得 $x_{4} = - \lambda_{2}x_{2} + \dfrac{1+ \lambda_{2}}{t}x_{1},y_{4} = - \lambda_{2}y_{2} + \dfrac{1+ \lambda_{2}}{t}y_{1}$，将其代入曲线方程可得$$\lambda_{2}^2(ax_{2}^2 + by_{2}^2) - \dfrac{2\lambda_{2}}{t}(ax_{1}x_{2} + by_{1}y_{2}) + \dfrac{(1+\lambda_{2})^2}{t^2}(ax_{1}^2 + by_{1}^2)=1,$$即 $\lambda_{2}^2 +\dfrac{(1+\lambda_{2})^2}{t^2}=1$，得 $\lambda_{2} = \dfrac{t^2 - 1}{t^2 +1}$，则 $\lambda_{1}=\lambda_{2}$ 得证．
故 $\dfrac{ON}{OA} =\dfrac{NP}{PM}$ 等价于 $\dfrac{NB}{MA}=\dfrac{ND}{MC} $ 等价于 $\dfrac{ND}{NB}=\dfrac{MC}{MA}.$
设 $\dfrac{MC}{MA} = \lambda _{1}$，$\dfrac{ND}{NB}=\lambda_{2}$，问题等价于证明 $\lambda _{1}=\lambda _{2}$．
设 $A(x_{1},y_{1})$、$B(x_{2},y_{2})$、$C(x_{3},y_{3})$、$D(x_{4},y_{4})$，则由 $k_{l_{1}} \cdot k_{l_{2}} = -\dfrac{a}{b}$，得 $ax_{1}x_{2} + by_{1}y_{2}=0$．由于 $\dfrac{OM}{OB}=\dfrac{OA}{ON} $，故可设 $M(tx_{2},ty_{2})$、$N(\dfrac{x_{1}}{t},\dfrac{y_{1}}{t})$，由 $\dfrac{MC}{MA} = \lambda _{1}$，可得 $\overrightarrow {MC} = \lambda _{1}\overrightarrow {MA}.$
即 $(x_{3} - tx_{2},y_{3} - ty_{2}) = \lambda _{1}(x_{1} - tx_{2},y_{1} - ty_{2})$，则$$x_{3} = \lambda _{1}x_{1} + (1-\lambda _{1})tx_{2},y_{3} =\lambda _{1}y_{1} + (1-\lambda _{1})ty_{2}.$$将 $C(x_{3},y_{3})$ 代入椭圆：$ax^2 +by^2 = 1$ 方程可得$$\lambda_{1}^2(ax_{1}^2 + by_{1}^2) + 2\lambda_{1}(1-\lambda_{1})t(ax_{1}x_{2} + by_{1}y_{2}) + (1-\lambda_{1})^2 t^2(ax_{2}^2 +by_{2}^2)=1,$$即 $\lambda_{1}^2 + (1-\lambda_{1})^2t^2 = 1$，得 $\lambda_{1} = \dfrac{t^2 - 1}{t^2 + 1} .$
同理由 $\dfrac{ND}{NB}=\lambda_{2}$，可得 $\overrightarrow {ND} = -\lambda_{2}\overrightarrow {NB}$，即$$(x_{4} - \dfrac{x_{1}}{t},y_{4} - \dfrac{y_{1}}{t})=-\lambda_{2}(x_{2} - \dfrac{x_{1}}{t},y_{2} - \dfrac{y_{1}}{t}).$$整理可得 $x_{4} = - \lambda_{2}x_{2} + \dfrac{1+ \lambda_{2}}{t}x_{1},y_{4} = - \lambda_{2}y_{2} + \dfrac{1+ \lambda_{2}}{t}y_{1}$，将其代入曲线方程可得$$\lambda_{2}^2(ax_{2}^2 + by_{2}^2) - \dfrac{2\lambda_{2}}{t}(ax_{1}x_{2} + by_{1}y_{2}) + \dfrac{(1+\lambda_{2})^2}{t^2}(ax_{1}^2 + by_{1}^2)=1,$$即 $\lambda_{2}^2 +\dfrac{(1+\lambda_{2})^2}{t^2}=1$，得 $\lambda_{2} = \dfrac{t^2 - 1}{t^2 +1}$，则 $\lambda_{1}=\lambda_{2}$ 得证．
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