每日一题[677]椭圆的“垂径定理”

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．

---

分析与解　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 $$\begin{split} \dfrac{d}{|MN|}&=\dfrac{\left|\dfrac{b^2m^2}{a^2n}+n\right|}{\sqrt{1+\dfrac{b^4m^2}{a^4n^2}}}\cdot \dfrac{1}{\sqrt{1+\dfrac{a^4n^2}{b^4m^2}}\cdot |m-s|}\\&=\dfrac{\left(\dfrac{b^2m^2}{a^2n^2}+1\right)\left(\dfrac{a^2n}{b^2m}+\dfrac{b^4m}{a^4n}\right)}{\sqrt{\left(1+\dfrac{b^4m^2}{a^4n^2}\right)\left(1+\dfrac{a^4n^2}{b^4m^2}\right)}\cdot\left(\dfrac{b^4m^2}{a^4n^2}+1\right)}\\&=\dfrac{\left(\dfrac{b^2m^2}{a^2n^2}+1\right)\left(1+\dfrac{b^6m^2}{a^6n^2}\right)}{\left(\dfrac{b^4m^2}{a^4n^2}+1\right)^2} ,\end{split}$$ 记$\dfrac{b^2m}{a^2n}=u$($u>0$)，$\dfrac{a}{b}=\lambda$，则 $$\begin{split} \dfrac{d}{|MN|}&=\dfrac{\left(\lambda ^2u^2+1\right)\left(1+\dfrac{u^2}{\lambda^2}\right)}{\left(u^2+1\right)^2}\\&=\dfrac{u^4+\lambda^2u^2+\dfrac{u^2}{\lambda^2}+1}{u^4+2u^2+1}\\&=1+\dfrac{\lambda^2+\dfrac{1}{\lambda^2}-2}{u^2+\dfrac{1}{u^2}+2}\\&\leqslant 1+\dfrac{\lambda^2+\dfrac{1}{\lambda^2}-2}4 ,\end{split}$$ 等号当且仅当$u=1$时取得．因此所求的最大值为$\dfrac 14\left(\dfrac ab+\dfrac ba\right)^2$．