过椭圆 $\dfrac{x^{2}}{18}+\dfrac{y^{2}}{9}=1$ 焦点的两条直线互相垂直，且 $A,B,C,D$ 均在椭圆上，求 $|AB|+|CD|$ 的最小值．
【解析】
弦长公式椭圆方程为 $\dfrac{x^{2}}{18}+\dfrac{y^{2}}{9}=1$，设直线方程为 $x=my+3$，则联立椭圆方程与直线方程求弦长：\[\begin{cases}x^{2}+2y^{2}-18=0,\\ x=my+3.\end{cases}\]即 $(m^{2}+2)y^{2}+6my-9=0$，于是弦长\[\sqrt{1+m^{2}}|y_{1}-y_{2}|=\sqrt{1+m^{2}}\cdot \dfrac{\sqrt{36m^{2}+36(m^{2}+2)}}{m^{2}+2}=\dfrac{6\sqrt 2(m^{2}+1)}{m^{2}+2}.\]因此若设直线 $AB$ 斜率的倒数为 $m$，则\[|AB|+|CD|=\dfrac{6\sqrt 2(m^{2}+1)}{m^{2}+2}+\dfrac{6\sqrt 2\left(\dfrac{1}{m^{2}}+1\right)}{\left(\dfrac{1}{m^{2}}+2\right)}=6\sqrt 2\left(\dfrac{m^{2}+1}{m^{2}+2}+\dfrac{m^{2}+1}{2m^{2}+1}\right)\geqslant 8\sqrt 2.\]当且仅当 $m^{2}=1$ 时取得等号．于是 $|AB|+|CD|$ 的最小值为 $8\sqrt 2$．
焦点弦长公式设直线 $AB$、$CD$ 的倾斜角分别为 $\alpha$、$\alpha+\dfrac{\pi}{2}$，则\[\begin{split}|AB|+|CD|&=\dfrac{2ab^{2}}{b^{2}+c^{2}\sin^{2}\alpha}+\dfrac{2ab^{2}}{b^{2}+c^{2}\cos^{2}\alpha}\\&=\dfrac{2ab^{2}\left[\left(b^{2}+c^{2}\cos^{2}\alpha\right)+\left(b^{2}+c^{2}\sin^{2}\alpha\right)\right]}{\left(b^{2}+c^{2}\sin^{2}\alpha\right)\left(b^{2}+c^{2}\cos^{2}\alpha\right)}\\&=\dfrac{2ab^{2}\left(2b^{2}+c^{2}\right)}{b^{4}+b^{2}c^{2}+c^{4}\left(\sin\alpha+\cos\alpha\right)^{2}}\\&=\dfrac{2ab^{2}\left(2b^{2}+c^{2}\right)}{b^{4}+b^{2}c^{2}+\dfrac{1}{4}c^{4}\sin^{2}2\alpha}.\end{split}\]所以当 $\alpha=\dfrac{\pi}{4}$ 时，$|AB|+|CD|$ 取得最小值为\[\dfrac{8ab^{2}}{2b^{2}+c^{2}}=\dfrac{8\cdot 3\sqrt 2\cdot 3^{2}}{2\cdot 3^{2}+3^{2}}=8\sqrt 2.\]
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