1694

已知曲线 ${\displaystyle E:\frac{x^2}{4}+\frac{y^2}{b^2}=1}$ (${\displaystyle b>0}$) 与点 ${\displaystyle P(\sqrt{5},0)}$, ${\displaystyle O}$ 为原点, 动点 ${\displaystyle Q\in E}$, 且 ${\displaystyle \mathrm{\angle }OPQ}$ 的最大值为 ${\displaystyle \frac{\pi }{4}}$.

1. 求曲线 ${\displaystyle E}$ 的方程;
2. 已知有 ${\displaystyle n+1}$ 个点 ${\displaystyle A_0,A_1,A_2,\cdots ,A_n}$ 按逆时针顺序依次在 ${\displaystyle E}$ 上, 且 ${\displaystyle A_0(2,0)}$, ${\displaystyle A_n(-2,0)}$.
   1. 当 ${\displaystyle A_1,A_2}$ 关于 ${\displaystyle y}$ 轴对称, 且 ${\displaystyle \mathrm{\Delta }O{A}_1{A}_3}$ 的面积为 ${\displaystyle 1}$ 时, 求直线 ${\displaystyle A_2{A}_3}$ 的斜率;
   2. 当 ${\displaystyle \mathrm{\Delta }O{A}_{k-1}{A}_k}$ (${\displaystyle 1\le k\le n}$) 的面积都相等时, 记多边形 ${\displaystyle A_0{A}_1{A}_2\cdots A_n}$ 的周长为 ${\displaystyle C_n}$. 若对于 ${\displaystyle \mathrm{\forall }n\in {\mathbb{N}}^{\ast }}$, 都有 ${\displaystyle C_n<\lambda }$, 求整数 ${\displaystyle \lambda }$ 的最小值.

1. 设 ${\displaystyle Q(2\cos \theta ,b\sin \theta )}$, 则 ${\displaystyle \mathrm{\angle }OPQ}$ 的最大值为 ${\displaystyle \frac{\pi }{4}}$ 意味着 ${\displaystyle |k_{PQ}|=|\tan \mathrm{\angle }OPQ|}$ 有最大值 ${\displaystyle 1}$. 也即 $$\left|\frac{b\sin \theta }{\sqrt{5}-2\cos \theta }\right|\le \frac{b|\sin \theta |}{|(2\cos \theta +\sin \theta )-2\cos \theta |}=b=1,$$ 所以 ${\displaystyle E:\frac{x^2}{4}+y^2=1}$.
2. 1. 设 ${\displaystyle A_1(2\cos \alpha ,\sin \alpha )}$, ${\displaystyle A_3(2\cos \beta ,\sin \beta )}$, 且 ${\displaystyle A_2(-2\cos \alpha ,\sin \alpha )}$. 因为 ${\displaystyle A_1,A_2,A_3}$ 是逆时针的位置关系, 则 ${\displaystyle 0<\alpha <\frac{\pi }{2}<\beta <\pi }$. 于是 $$S_{\mathrm{\Delta }O{A}_1{A}_3}=\frac{1}{2}|\overrightarrow{O{A}_1}\times \overrightarrow{O{A}_3}|=|\cos \beta \sin \alpha -\cos \alpha \sin \beta |=|\sin (\alpha -\beta )|=1,$$ 于是只有 ${\displaystyle \alpha -\beta =-\frac{\pi }{2}}$. 这样 $$k_{A_2{A}_3}=\frac{\sin \alpha -\sin \beta }{-2(\cos \alpha +\cos \beta )}=\frac{\sin \alpha -\cos \alpha }{-2(\cos \alpha -\sin \alpha )}=\frac{1}{2}.$$
   2. 设 ${\displaystyle A_k(2\cos {\theta }_k,\sin {\theta }_k)}$, 则类似 2.1, 有 ${\displaystyle S_{\mathrm{\Delta }O{A}_{k-1}{A}_k}=\sin ({\theta }_k-{\theta }_{k-1})}$ 对任意 ${\displaystyle k}$ 都相等, 也即 ${\displaystyle {\theta }_k-{\theta }_{k-1}}$ 对任意 ${\displaystyle k}$ 都相等. 而 ${\displaystyle {\theta }_0=0}$, ${\displaystyle {\theta }_n=\pi }$, 所以 ${\displaystyle \{{\theta }_k{\}}_{k=1}^n}$ 是等差数列, ${\displaystyle {\theta }_k=\frac{k\pi }{n}}$. 于是 $$\begin{array}{rl}|A_{k-1}{A}_k{|}^2& =(2\cos {\theta }_{k-1}-2\cos {\theta }_k{)}^2+(\sin {\theta }_{k-1}-\sin {\theta }_k{)}^2\\ & =16{\sin }^2\frac{{\theta }_{k-1}+{\theta }_k}{2}{\sin }^2\frac{{\theta }_{k-1}-{\theta }_k}{2}+4{\cos }^2\frac{{\theta }_{k-1}+{\theta }_k}{2}{\sin }^2\frac{{\theta }_{k-1}-{\theta }_k}{2}\\ & =2{\sin }^2\frac{{\theta }_{k-1}-{\theta }_k}{2}(2+6{\sin }^2\frac{{\theta }_{k-1}+{\theta }_k}{2})\\ & =2{\sin }^2\frac{\pi }{2n}[5-3\cos ({\theta }_{k-1}+{\theta }_k)]\\ & =2{\sin }^2\frac{\pi }{2n}(5-3\cos \frac{(2k-1)\pi }{n}).\end{array}$$
      此外我们说明 $$\begin{array}{rl}\sum _{k=1}^n\cos \frac{(2k-1)\pi }{n}& =\frac{1}{\sin \frac{\pi }{n}}\sum _{k=1}^n\sin \frac{\pi }{n}\cos \frac{(2k-1)\pi }{n}\\ & =\frac{1}{\sin \frac{\pi }{n}}\sum _{k=1}^n[\sin (\frac{(2k-1)\pi }{n}+\frac{\pi }{n})-\sin (\frac{(2k-1)\pi }{n}-\frac{\pi }{n})]\\ & =\frac{1}{\sin \frac{\pi }{n}}\sum _{k=1}^n[\sin \frac{2k\pi }{n}-\sin \frac{(2k-2)\pi }{n}]\\ & =\frac{1}{\sin \frac{\pi }{n}}(\sin 2\pi -\sin 0)=0.\end{array}$$ 于是$$\begin{array}{rl}{C}_n& =4+\sum _{k=1}^n|A_{k-1}{A}_k|\le 4+\sqrt{n\sum _{k=1}^n|A_{k-1}{A}_k{|}^2}\\ & =4+\sqrt{n\cdot 10n{\sin }^2\frac{\pi }{2n}}<4+\sqrt{10n^2{\left(\frac{\pi }{2n}\right)}^2}=4+\frac{\pi \sqrt{10}}{2}.\end{array}$$ 而 ${\displaystyle {\pi }^2<{3.15}^2<10}$, 也即 ${\displaystyle \frac{\pi \sqrt{10}}{2}<5}$, 于是 ${\displaystyle C_n<9}$. 另一方面如果 ${\displaystyle n=2}$, 则 ${\displaystyle C_n=4+\sqrt{20}>8}$, 所以 ${\displaystyle \lambda }$ 的最小值为 ${\displaystyle 9}$.