已知椭圆 $E:\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$（$a>b>0$），$M$ 为椭圆内不在坐标轴上一点．过 $M$ 作不过原点的直线交椭圆于 $A,B$ 两点，$M$ 恰为 $AB$ 的中点，过 $M$ 作 $AB$ 的垂线交椭圆于 $C,D$ 两点，$N$ 为弦 $CD$ 的中点．记 $O$ 到直线 $AB$ 的距离为 $d$，求 $\dfrac{d}{|MN|}$ 的最大值．
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【答案】
$\dfrac 14\left(\dfrac ab+\dfrac ba\right)^2$
【解析】
不妨设 $M(m,n)$，$m,n>0$，则根据椭圆的"垂径定理"，可得直线 $AB$ 的斜率为 $-\dfrac{b^2m}{a^2n}$，于是$$AB:y=-\dfrac{b^2m}{a^2n}(x-m)+n,$$进而由 $AB\perp CD$，可得$$CD:y=\dfrac{a^2n}{b^2m}(x-m)+n.$$设 $N(s,t)$，则由椭圆的"垂径定理"，有$$t=-\dfrac{b^4m}{a^4n}s,$$又$$t=\dfrac{a^2n}{b^2m}(s-m)+n,$$于是$$s=\dfrac{\dfrac{a^2n}{b^2}-n}{\dfrac{a^2n}{b^2m}+\dfrac{b^4m}{a^4n}},$$因此\begin{eqnarray*}\begin{split} \dfrac{d}{|MN|}&=\dfrac{\left|\dfrac{b^2m^2}{a^2n}+n\right|}{\sqrt{1+\dfrac{b^4m^2}{a^4n^2}}}\cdot \dfrac{1}{\sqrt{1+\dfrac{a^4n^2}{b^4m^2}}\cdot |m-s|}\\
&=\dfrac{\left(\dfrac{b^2m^2}{a^2n^2}+1\right)\left(\dfrac{a^2n}{b^2m}+\dfrac{b^4m}{a^4n}\right)}{\sqrt{\left(1+\dfrac{b^4m^2}{a^4n^2}\right)\left(1+\dfrac{a^4n^2}{b^4m^2}\right)}\cdot\left(\dfrac{b^4m^2}{a^4n^2}+1\right)}\\
&=\dfrac{\left(\dfrac{b^2m^2}{a^2n^2}+1\right)\left(1+\dfrac{b^6m^2}{a^6n^2}\right)}{\left(\dfrac{b^4m^2}{a^4n^2}+1\right)^2}
,\end{split} \end{eqnarray*}记 $\dfrac{b^2m}{a^2n}=u$（$u>0$），$\dfrac{a}{b}=\lambda$，则\begin{eqnarray*}\begin{split} \dfrac{d}{|MN|}&=\dfrac{\left(\lambda ^2u^2+1\right)\left(1+\dfrac{u^2}{\lambda^2}\right)}{\left(u^2+1\right)^2}\\
&=\dfrac{u^4+\lambda^2u^2+\dfrac{u^2}{\lambda^2}+1}{u^4+2u^2+1}\\
&=1+\dfrac{\lambda^2+\dfrac{1}{\lambda^2}-2}{u^2+\dfrac{1}{u^2}+2}\\
&\leqslant 1+\dfrac{\lambda^2+\dfrac{1}{\lambda^2}-2}4
,\end{split} \end{eqnarray*}等号当且仅当 $u=1$ 时取得．因此所求的最大值为 $\dfrac 14\left(\dfrac ab+\dfrac ba\right)^2$．
&=\dfrac{\left(\dfrac{b^2m^2}{a^2n^2}+1\right)\left(\dfrac{a^2n}{b^2m}+\dfrac{b^4m}{a^4n}\right)}{\sqrt{\left(1+\dfrac{b^4m^2}{a^4n^2}\right)\left(1+\dfrac{a^4n^2}{b^4m^2}\right)}\cdot\left(\dfrac{b^4m^2}{a^4n^2}+1\right)}\\
&=\dfrac{\left(\dfrac{b^2m^2}{a^2n^2}+1\right)\left(1+\dfrac{b^6m^2}{a^6n^2}\right)}{\left(\dfrac{b^4m^2}{a^4n^2}+1\right)^2}
,\end{split} \end{eqnarray*}记 $\dfrac{b^2m}{a^2n}=u$（$u>0$），$\dfrac{a}{b}=\lambda$，则\begin{eqnarray*}\begin{split} \dfrac{d}{|MN|}&=\dfrac{\left(\lambda ^2u^2+1\right)\left(1+\dfrac{u^2}{\lambda^2}\right)}{\left(u^2+1\right)^2}\\
&=\dfrac{u^4+\lambda^2u^2+\dfrac{u^2}{\lambda^2}+1}{u^4+2u^2+1}\\
&=1+\dfrac{\lambda^2+\dfrac{1}{\lambda^2}-2}{u^2+\dfrac{1}{u^2}+2}\\
&\leqslant 1+\dfrac{\lambda^2+\dfrac{1}{\lambda^2}-2}4
,\end{split} \end{eqnarray*}等号当且仅当 $u=1$ 时取得．因此所求的最大值为 $\dfrac 14\left(\dfrac ab+\dfrac ba\right)^2$．
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