一道解析几何题的三种解法

已知椭圆$E:\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$($a>b>0$)所在平面内有一个不与原点重合的点$P(x_0,y_0)$，过$P$作$E$的任意两条割线$AB,CD$，其中$A,B,C,D$均在椭圆$E$上．证明：直线$AC$和$BD$的交点在定直线上．

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分析与解　如图，设直线$AC$与$BD$的交点为$Q$，直线$AD$与$BC$的交点为$R$，则直线$RQ$即题中的定直线．

法一　几何方法
如图，设直线$AB,CD$分别与直线$RQ$相交于$S,T$，则利用完全四边形$ABCD$可得四点$S,R,T,Q$为共线的调和点列．下面证明直线$SRTQ$由$P$点位置确定，与直线$AB$和$CD$的作法无关．
在$A,B$处作切线，两条切线交于$M$；在$C,D$处作切线，两条切线交于$N$．根据帕斯卡(Pascal)定理，$M,N$均在直线$ST$上，即$S,R,T,Q,M,N$六点共线．但从图中可以看出，$S,M$两点只和直线$AB$有关，与直线$CD$完全无关，所以我们将直线$AB$换成任意的另一条直线$A'B'$，只要$A'B'$通过点$P$，则形成的$S,Q,T,Q$四点仍与$M,N$共线．类似的，可以证明直线$SRTQ$的位置与直线$AB$完全无关，而只取决于$P$点位置，因此原命题得证．

注　帕斯卡(Pascal)定理：如果一个四边形内接于某圆锥曲线，则四边形的两组对边的两个交点与两组对顶点的切线的两个交点四点共线．

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法二　混合方法

首先给出引理．

引理　过点$P$的直线$l$与椭圆交于$S,T$两点，$X$为直线$ST$上的点，那么$P,X$调和分割$S,T$，即“$\dfrac{|SX|}{|XT|}=\dfrac{|SP|}{|PT|}$
”与“点$X$在直线$\dfrac{x_0x}{a^2}+\dfrac{y_0y}{b^2}=1$上”等价．

该引理利用直线的参数方程或定比点差法容易证明，此处从略．

根据引理，我们只需要证明调和分割$A,B$的点以及调和分割$C,D$的点在直线$RQ$上即可(完全四边形的性质)．设$AB,CD$与直线$RQ$的交点分别为$S,T$．事实上，对$\triangle CDQ$和截线$APB$应用梅涅劳斯定理，有

$$\dfrac{|DB|}{|BQ|}\cdot \dfrac{|QA|}{|AC|}\cdot \dfrac{|CP|}{|PD|}=1,$$ 对于$\triangle CDQ$和点$R$应用塞瓦定理，有 $$\dfrac{|DB|}{|BQ|}\cdot \dfrac{|QA|}{|AC|}\cdot \dfrac{|CT|}{|TD|}=1,$$ 于是$\dfrac{|CP|}{|PD|}=\dfrac{|CT|}{|TD|}$，因此$P,T$调和分割$C,D$．类似的，有$P,S$调和分割$A,B$．因此原命题得证．

法三　解析方法
设$AB$与$CD$，$AC$与$BD$的交点分别为$P(x_0,y_0)$，$Q(x_1,y_1)$，只需要证明$\dfrac{x_0x_1}{a^2}+\dfrac{y_0y_1}{b^2}=1$即可．设$A\left(a\cos\alpha,b\sin\alpha\right)$，$B\left(a\cos\beta,b\sin\beta\right)$，$C\left(a\cos\gamma,b\sin\gamma\right)$，$D\left(a\cos\delta,b\sin\delta\right)$，则

$$\begin{split} &AB:\dfrac xa\cdot \cos\dfrac{\alpha+\beta}2+\dfrac yb\cdot \sin\dfrac{\alpha+\beta}2=\cos\dfrac{\alpha-\beta}2,\\ &CD:\dfrac xa\cdot \cos\dfrac{\gamma+\delta}2+\dfrac yb\cdot \sin\dfrac{\gamma+\delta}2=\cos\dfrac{\gamma-\delta}2,\end{split}$$ 于是直线$AB$与$CD$的交点坐标为 $$\left(a\cdot \dfrac{\cos\dfrac{\alpha-\beta}2\sin\dfrac{\gamma+\delta}2-\sin\dfrac{\alpha+\beta}2\cos\dfrac{\gamma-\delta}2}{\cos\dfrac{\alpha+\beta}2\sin\dfrac{\gamma+\delta}2-\sin\dfrac{\alpha+\beta}2\cos\dfrac{\gamma+\delta}2},b\cdot \dfrac{\cos\dfrac{\alpha-\beta}2\cos\dfrac{\gamma+\delta}2-\cos\dfrac{\alpha+\beta}2\cos\dfrac{\gamma-\delta}2}{\sin\dfrac{\alpha+\beta}2\cos\dfrac{\gamma+\delta}2-\cos\dfrac{\alpha+\beta}2\sin\dfrac{\gamma+\delta}2}\right)$$ 即 $$\left(a\cdot \dfrac{\sin\dfrac{\alpha+\beta}2\cos\dfrac{\gamma-\delta}2-\cos\dfrac{\alpha-\beta}2\sin\dfrac{\gamma +\delta}2}{\sin\dfrac{\alpha+\beta-\gamma-\delta}2},b\cdot\dfrac{\cos\dfrac{\alpha-\beta}2\cos\dfrac{\gamma+\delta}2-\cos\dfrac{\alpha+\beta}2\cos\dfrac{\gamma-\delta}2}{\sin\dfrac{\alpha+\beta-\gamma-\delta}2}\right),$$ 交换$\gamma$和$\beta$就可以得到$AC$与$BD$的交点坐标，为 $$\left(a\cdot \dfrac{\sin\dfrac{\alpha+\gamma}2\cos\dfrac{\beta-\delta}2-\cos\dfrac{\alpha-\gamma}2\sin\dfrac{\beta +\delta}2}{\sin\dfrac{\alpha+\gamma-\beta-\delta}2},b\cdot\dfrac{\cos\dfrac{\alpha-\gamma}2\cos\dfrac{\beta+\delta}2-\cos\dfrac{\alpha+\gamma}2\cos\dfrac{\beta-\delta}2}{\sin\dfrac{\alpha+\gamma-\beta-\delta}2}\right).$$ 于是我们只需证明 $$\begin{split}& \left(\sin\dfrac{\alpha+\beta}2\cos\dfrac{\gamma-\delta}2-\cos\dfrac{\alpha-\beta}2\sin\dfrac{\gamma +\delta}2\right)\cdot\left(\sin\dfrac{\alpha+\gamma}2\cos\dfrac{\beta-\delta}2-\cos\dfrac{\alpha-\gamma}2\sin\dfrac{\beta +\delta}2\right) \\ &\qquad\qquad +\left(\cos\dfrac{\alpha-\beta}2\cos\dfrac{\gamma+\delta}2-\cos\dfrac{\alpha+\beta}2\cos\dfrac{\gamma-\delta}2\right)\cdot \left(\cos\dfrac{\alpha-\gamma}2\cos\dfrac{\beta+\delta}2-\cos\dfrac{\alpha+\gamma}2\cos\dfrac{\beta-\delta}2\right)\\ &\qquad\qquad\qquad\qquad\qquad \qquad\qquad \qquad\qquad \qquad\qquad \qquad\qquad =\sin\dfrac{\alpha+\beta-\gamma-\delta}2\cdot \sin\dfrac{\alpha+\gamma-\beta-\delta}2.\end{split}$$ 事实上，有 $$\begin{split} LHS&=\cos\dfrac{\gamma-\delta}2\cos\dfrac{\beta-\delta}2\left(\sin\dfrac{\alpha+\beta}2\sin\dfrac{\alpha+\gamma}2+\cos\dfrac{\alpha+\beta}2\cos\dfrac{\alpha+\gamma}2\right)\\ &\qquad \qquad +\cos\dfrac{\gamma-\delta}2\cos\dfrac{\alpha-\gamma}2\left(-\sin\dfrac{\alpha+\beta}2\sin\dfrac{\beta+\delta}2-\cos\dfrac{\alpha+\beta}2\cos\dfrac{\beta+\delta}2\right)\\ &\qquad \qquad+\cos\dfrac{\alpha-\beta}2\cos\dfrac{\beta-\delta}2\left(-\sin\dfrac {\gamma+\delta}2\sin\dfrac{\alpha+\gamma}2-\cos\dfrac {\gamma+\delta}2\cos\dfrac{\alpha+\gamma}2\right)\\ &\qquad \qquad+\cos\dfrac{\alpha-\beta}2\cos\dfrac{\alpha-\gamma}2\left(\sin\dfrac{\gamma+\delta}2\sin\dfrac{\beta+\delta}2+\cos\dfrac{\gamma+\delta}2\cos\dfrac{\beta+\delta}2\right)\\ &=\cos\dfrac{\gamma-\delta}2\cos\dfrac{\beta-\delta}2\cos\dfrac{\beta-\gamma}2-\cos\dfrac{\gamma-\delta}2\cos\dfrac{\alpha-\gamma}2\cos\dfrac{\alpha-\delta}2\\ &\qquad \qquad-\cos\dfrac{\alpha-\beta}2\cos\dfrac{\beta-\delta}2\cos\dfrac{\alpha-\delta}2+\cos\dfrac{\alpha-\beta}2\cos\dfrac{\alpha-\gamma}2\cos\dfrac{\beta-\gamma}2\\ &=\cos\dfrac{\gamma-\delta }2\cos\left(\dfrac{\alpha-\beta}2+\dfrac{\delta-\alpha}2\right)\cos\dfrac{\beta-\gamma}2-\cos\dfrac{\gamma-\delta}2\cos\left(\dfrac{\alpha-\beta}2+\dfrac{\beta-\gamma}2\right)\cos\dfrac{\delta-\alpha}2\\ &\qquad \qquad-\cos\dfrac{\alpha-\beta}2\cos\left(\dfrac{\beta-\gamma}2+ \dfrac{\gamma-\delta}2\right)\cos\dfrac{\delta-\alpha}2+\cos\dfrac{\alpha-\beta}2\cos\left(\dfrac{\gamma-\delta}2+\dfrac{\delta-\alpha}2\right)\cos\dfrac{\beta-\gamma}2\\ &=\sin\dfrac{\alpha-\beta}2\sin\dfrac{\beta-\gamma}2\cos\dfrac{\gamma-\delta}2\cos\dfrac{\delta-\alpha}2-\sin\dfrac{\alpha-\beta}2\cos\dfrac{\beta-\gamma}2\cos\dfrac{\gamma-\delta}2\sin\dfrac{\delta-\alpha}2\\ &\qquad \qquad+\cos\dfrac{\alpha-\beta}2\sin\dfrac{\beta-\gamma}2\sin\dfrac{\gamma-\delta}2\cos\dfrac{\delta-\alpha}2-\cos\dfrac{\alpha-\beta}2\cos\dfrac{\beta-\gamma}2\sin\dfrac{\gamma-\delta}2\sin\dfrac{\delta-\alpha}2\\ &=\sin\dfrac{\alpha-\beta}2\cos\dfrac{\gamma-\delta}2\left(\sin\dfrac{\beta-\gamma}2\cos\dfrac{\delta-\alpha}2-\cos\dfrac{\beta-\gamma}2\sin\dfrac{\delta -\alpha}2\right)\\ &\qquad \qquad+\cos\dfrac{\alpha-\beta}2\sin\dfrac{\gamma-\delta}2\left(\sin\dfrac{\beta-\gamma}2\cos\dfrac{\delta-\alpha}2-\cos\dfrac{\beta-\gamma}2\sin\dfrac{\delta-\alpha}2\right)\\ &=\sin\dfrac{\alpha-\beta}2\cos\dfrac{\gamma-\delta}2\sin\dfrac{\alpha+\beta-\gamma-\delta}2+\cos\dfrac{\alpha-\beta}2\sin\dfrac{\gamma-\delta}2\sin\dfrac{\alpha+\beta-\gamma-\delta}2\\ &=\sin\dfrac{\alpha-\beta+\gamma-\delta}2\sin\dfrac{\alpha+\beta-\gamma-\delta}2\\ &=RHS,\end{split}$$ 因此原命题得证．