已知椭圆$E:\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$($a>b>0$)，$A$为椭圆$E$的右顶点，$M,N$是椭圆$E$上不同于$A$的不同两点，且直线$AM$和$AN$的斜率之积为$\lambda $．
(1) 求证：直线$MN$过定点$R$；
(2) 若$\lambda =-\dfrac{b^2}{a^2}$，$P$为椭圆$E$上不同于$M,N$的一点，且$|PM|=|PN|$，求$\triangle MNP$的面积的最小值．
分析与解 (1) 平移坐标系$xOy$至以$A$为坐标原点，则$E':\dfrac{(x'+a)^2}{a^2}+\dfrac{y'^2}{b^2}=1$，即\[E':\dfrac{x'^2}{a^2}+\dfrac{y'^2}{b^2}+\dfrac{2}{a}x'=0,\]设直线$M'N':mx'+ny'=1$，与椭圆$E'$的方程化齐次联立，得\[\dfrac{x'^2}{a^2}+\dfrac{y'^2}{b^2}+\dfrac{2}{a}x'(mx'+ny')=0,\]即\[\dfrac{1}{b^2}y'^2+\dfrac{2n}{a}x'y'+\left(\dfrac{1}{a^2}+\dfrac{2m}a\right)x'^2=0,\]于是由直线$A'M'$与直线$A'N'$斜率之积为$\lambda $，可得\[\dfrac{\dfrac{1}{a^2}+\dfrac{2m}a}{\dfrac{1}{b^2}}=\lambda ,\]因此\[\dfrac{1}{m}=\dfrac{2ab^2}{\lambda a^2-b^2},\]为定值．进而直线$MN$过定点$R'\left(\dfrac{2ab^2}{\lambda a^2-b^2},0\right)$，回到原坐标，定点为$R\left(\dfrac{\lambda a^2+b^2}{\lambda a^2-b^2}\cdot a,0\right)$．
(2) 当$\lambda =-\dfrac {b^2}{a^2}$时，$R$即坐标原点$O$．此时$O$点平分线段$MN$，因此$OP\perp MN$．不妨设$M\left(r_1\cos\theta,r_1\sin\theta\right)$，$P\left(r_2\cos\left(\theta+\dfrac{\pi}2\right),r_2\sin\left(\theta+\dfrac{\pi}2\right)\right)$．此时有\[\begin{aligned}\dfrac{(r_1\cos\theta)^2}{a^2}+\dfrac {(r_1\sin\theta)^2}{b^2}=1,\\\dfrac{\left(r_2\cos\left(\theta+\dfrac{\pi}2\right)\right)^2}{a^2}+\dfrac{\left(r_2\sin\left(\theta+\dfrac{\pi}2\right)\right)^2}{b^2}=1,
\end{aligned} \]因此\[\dfrac{1}{r_1^2}+\dfrac{1}{r_2^2}=\dfrac{1}{a^2}+\dfrac{1}{b^2}\,\]而$\triangle PMN$的面积为\[\begin{split}\dfrac 12 |MN|\cdot |OP|&= r_1r_2\\&=\dfrac{r_1r_2\cdot \left(\dfrac{1}{r_1^2}+\dfrac{1}{r_2^2}\right)}{\dfrac{1}{a^2}+\dfrac{1}{b^2}}\\&=\dfrac{\dfrac{r_2}{r_1}+\dfrac{r_1}{r_2}}{\dfrac{1}{a^2}+\dfrac{1}{b^2}}\\&\geqslant \dfrac {2a^2b^2}{a^2+b^2},\end{split}\]等号当$r_1=r_2$时(取$\theta=-\dfrac{\pi}4$)取得．因此所求的最小值为$\dfrac{2a^2b^2}{a^2+b^2}$．
注 注意到根据椭圆的「垂径定理」，椭圆$E$上的任意一点与关于原点对称的两点的连线的斜率之积(当斜率均存在时)为$-\dfrac{b^2}{a^2}$，因此若$\lambda =-\dfrac{b^2}{a^2}$，则直线$MN$恒过原点$O$．