题目一
题文
今有椭圆 \(\Gamma: \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1, (a > b > 0)\) 点 \(A, B \in \Gamma\) 并且 \(OA \bot OB\),求 \(O\) 到 \(AB\) 的射影 \(H\) 的轨迹方程。
解法一
考虑设
\[OA: y=kx, OB: y=-\frac{x}{k} \]先考察斜率存在的情形。联立 \(OA, \Gamma\) 得
\[\begin{cases} \Gamma :& b^2 x^2 + a^2 y^2 - a^2 b^2 = 0\\ OA :& y = kx \end{cases} \implies x_1^2 = \frac{a^2 b^2}{b^2 + a^2k^2} \]联立 \(OB, \Gamma\) 得
\[\begin{cases} \Gamma :& b^2 x^2 + a^2 y^2 - a^2 b^2 = 0 \\ OB :& y = -\frac{x}{k} \end{cases} \implies x_2^2 = \frac{a^2 b^2}{b^2 + \frac{a^2}{k^2}} \]即
\[OA^2 = \frac{(k^2 + 1) a^2 b^2}{b^2 + a^2k^2} \]\[OB^2 = \frac{(\frac{1}{k^2} + 1) a^2 b^2}{b^2 + \frac{a^2}{k^2}} \]根据面积关系
\[2S = OA \cdot OB = OH \cdot AB \implies OH^2 = \frac{OA^2 OB^2}{AB^2} = \frac{OA^2 OB^2}{OA^2 + OB^2} \]我们得到
\[\begin{equation*} \begin{split} \frac{1}{OH^2} &= \frac{1}{OA^2} + \frac{1}{OB^2}\\ &=\frac{b^2 + a^2 k^2}{(k^2 + 1) a^2 b^2} + \frac{b^2 + \frac{a^2}{k^2}}{(\frac{1}{k^2} + 1) a^2 b^2}\\ &=\frac{b^2 + a^2 k^2}{(k^2 + 1) a^2 b^2} + \frac{k^2 b^2 + a^2}{(k^2 + 1) a^2 b^2} = \frac{(k^2 + 1)(a^2 + b^2)}{(k^2 + 1)a^2 b^2} = \frac{a^2 + b^2}{a^2 b^2} \end{split} \end{equation*} \]即
\[OH^2 = \frac{a^2 b^2}{a^2 + b^2} \]再考察斜率不存在的情形(略)。容易发现点 \(H\) 的轨迹方程为
\[x^2 + y^2 = \frac{a^2 b^2}{a^2 + b^2} \]解法二
解法一的代数运算比较复杂。我们考虑用别的东西代替斜率。
设
\[OA = r_1, OB = r_2 \]则
\[\begin{equation*} \begin{split} \mathbf{OA}&=(r_1 \sin \alpha, r_1 \cos \alpha)\\ \mathbf{OB}&=\left(r_2 \sin (\alpha + \frac{\pi}{2}), r_2 \cos (\alpha + \frac{\pi}{2})\right) = (-r_2 \cos \alpha, r_2 \sin \alpha)\\ \end{split} \end{equation*} \]将 \(A, B\) 坐标代入 \(\Gamma\) 得到
\[\begin{cases} \displaystyle\frac{r_1^2 \sin^2 \alpha}{a^2} + \frac{r_1^2 \cos^2 \alpha}{b^2} = 1\\ \displaystyle\frac{r_2^2 \cos^2 \alpha}{a^2} + \frac{r_2^2 \sin^2 \alpha}{b^2} = 1 \end{cases} \]即
\[\begin{cases} \displaystyle\frac{\sin^2 \alpha}{a^2} + \frac{\cos^2 \alpha}{b^2} = \frac{1}{r_1^2}\\ \displaystyle\frac{\cos^2 \alpha}{a^2} + \frac{\sin^2 \alpha}{b^2} = \frac{1}{r_2^2} \end{cases} \]两式相加,得到
\[\frac{\sin^2 \alpha + \cos^2 \alpha}{a^2} + \frac{\sin^2 \alpha + \cos^2 \alpha}{b^2} = \frac{1}{a^2} + \frac{1}{b^2} = \frac{1}{r^2_1} + \frac{1}{r^2_2} \]而由面积关系
\[\frac{1}{OH^2} = \frac{1}{OA^2} + \frac{1}{OB^2} = \frac{1}{r_1^2} + \frac{1}{r_2^2} = \frac{1}{a^2} + \frac{1}{b^2}\\ \implies OH^2 = \frac{a^2 b^2}{a^2 + b^2} \implies x^2 + y^2 = \frac{a^2 b^2}{a^2 + b^2} \]题目二
题文
今有抛物线 \(W: y = x^2 + \frac{1}{4}\),又矩形 \(ABCD\) 恰有三个顶点在 \(W\) 上,证明 \(C_{ABCD} > 3 \sqrt{3}\)。
解法一
注意到 \(AB, BC\) 的斜率必然存在,否则其中一条与 \(W\) 仅有一个交点。
设
\[A(x_1, y_1), B(x_0, y_0), C(x_2, y_2) \]则
\[\begin{equation*} \begin{split} BA&: y = k(x - x_0) + x_0^2 + \frac{1}{4}\\ BC&: y = -\frac{1}{k}(x - x_0) + x_0^2 + \frac{1}{4} \end{split} \end{equation*} \]令 \(k \ge 0\)。联立 \(BA\) 与 \(W\) 得
\[\begin{cases} W:& y = x^2 + \frac{1}{4}\\ BA:& y = k(x - x_0) + x_0^2 + \frac{1}{4} \end{cases} \implies x^2 - kx + kx_0 - x_0^2 = 0 \]方程有根 \(x_0, x_1\),由韦达定理得
\[x_0 + x_1 = k \]联立 \(BC\) 与 \(W\) 得
\[\begin{cases} W:& y = x^2 + \frac{1}{4}\\ BC:& y = -\frac{1}{k}(x - x_0) + x_0^2 + \frac{1}{4} \end{cases} \implies x^2 + \frac{x}{k} - \frac{x_0}{k} - x_0^2 = 0 \]方程有根 \(x_0, x_2\),由韦达定理得
\[x_0 + x_2 = -\frac{1}{k} \]由 \(x_0, x_1, x_2\),矩形 \(ABCD\) 的周长
\[\begin{align*} C_{ABCD} &= 2\left(BA + BC\right) \\ &= 2 \sqrt{k^2 + 1} \cdot \left | x_1 - x_0 \right | + 2 \sqrt{\frac{1}{k^2} + 1} \cdot \left | x_2 - x_0 \right |\\ &= 2 \sqrt{k^2 + 1} \cdot \left | k - 2x_0 \right | + 2 \sqrt{\frac{1}{k^2} + 1} \cdot \left |-\frac{1}{k} - 2x_0 \right |\\ &= 2 \sqrt{k^2 + 1} \left( \left | 2x_0 - k \right | + \left | \frac{2}{k} x_0 + \frac{1}{k^2} \right | \right) \end{align*} \]由绝对值函数的性质,将 \(x_0\) 看作主元,最小值取到,当且仅当 \(x_0 = \frac{k}{2}\) 或 \(x_0 = -\frac{1}{2k}\)。即
\[\left(C_{ABCD}\right)_{\min} = \min\left\{ 2 \sqrt{k^2 + 1} \left | 1 + \frac{1}{k^2}\right |, 2 \sqrt{k^2 + 1} \left | k + \frac{1}{k} \right | \right\} \]记函数
\[f(k) = 2\sqrt{k^2 + 1} \left | 1 + \frac{1}{k^2} \right | \]即
\[(C_{ABCD})_{\min} = \min \left \{ f\left(k\right), f\left(\frac{1}{k}\right) \right \} \]以 \(\frac{1}{k'}\) 代 \(k\),发现 \(f(k) = f(\frac{1}{k'})\),则只须求 \(\min f(\frac{1}{k})\)。
注意到
\[\begin{equation*} \begin{split} f(\frac{1}{k}) &= 2 \sqrt{k^2 + 1} \cdot \frac{k^2 + 1}{k}\\ &= \frac{2(k^2 + 1)^{\frac{3}{2}}}{k} = \frac{2}{k} \left( k^2 + \frac{1}{2} + \frac{1}{2}\right)^{\frac{3}{2}}\\ &\ge \frac{2}{k} \left[ 3 \cdot \left( k^2 \cdot \frac{1}{2} \cdot \frac{1}{2} \right) ^{\frac{1}{3}} \right]^{\frac{3}{2}} = \frac{2 \cdot \frac{1}{2} \cdot 3^{\frac{3}{2}} \cdot k}{k} = 3\sqrt{3} \end{split} \end{equation*} \]取等当且仅当 \(k^2 = \frac{1}{2}\),即 \(k = \frac{\sqrt{2}}{2}\),但此时 \(x_0 = x_2 = -\frac{\sqrt{2}}{2}\),故取不到等号。证毕。
解法二
设
\[\begin{align*} \mathbf{BA} &= (r_1 \sin \alpha, r_1, \cos \alpha)\\ \mathbf{BC} &= (-r_2 \cos \alpha, r_2 \sin \alpha)\\ \mathbf{OB} &= (x_0, x_0^2 + \frac{1}{4}) \end{align*} \]则
\[\begin{align*} \mathbf{OA} &= \mathbf{OB} + \mathbf{BA}\\ &= (x_0 + r_1 \sin \alpha, x_0^2 + \frac{1}{4} + r_1 \cos \alpha)\\ \mathbf{OC} &= \mathbf{OB} + \mathbf{BC}\\ &= (x_0 - r_2 \cos \alpha, x_0^2 + \frac{1}{4} + r_2 \sin \alpha) \end{align*} \]将 \(A, C\) 坐标代入 \(W\) 得
\[\begin{cases} x_0^2 + \frac{1}{4} + r_1 \cos \alpha = (x_0 + r_1 \sin \alpha)^2 + \frac{1}{4}\\ x_0^2 + \frac{1}{4} + r_2 \sin \alpha = (x_0 - r_2 \cos \alpha)^2 + \frac{1}{4} \end{cases} \implies \begin{cases} \cos \alpha = 2 x_0 \sin \alpha + r_1 \sin^2 \alpha\\ \sin \alpha = - 2 x_0 \cos \alpha + r_2 \cos^2 \alpha \end{cases} \]整理得
\[\begin{cases} \cos^2 \alpha = 2 x_0 \sin \alpha \cos \alpha + r_1 \sin^2 \alpha \cos \alpha\\ \sin^2 \alpha = - 2 x_0 \cos \alpha \sin \alpha + r_2 \cos^2 \alpha \sin \alpha \end{cases} \]即
\[r_1 \sin^2 \alpha \cos \alpha + r_2 \cos^2 \alpha \sin \alpha = \sin^2 \alpha + \cos^2 \alpha = 1 \]不失一般性,设 \(\alpha \in \left[0, \frac{\pi}{2} \right)\)。再对其分类如下:
- \(\alpha \in \left[ \frac{\pi}{4}, \frac{\pi}{2} \right)\) 时,\(r_1 = \frac{1 - r_2 \cos^2 \alpha \sin \alpha}{\sin^2 \alpha \cos \alpha}\),则\[\begin{align*} r_1 + r_2 &= \frac{1}{\sin^2 \alpha \cos \alpha} + \frac{\sin \alpha - \cos \alpha}{\sin \alpha} r_2\\ &\ge \frac{1}{\sin^2 \alpha \cos \alpha} = \frac{1}{\cos \alpha - \cos^3 \alpha} \end{align*} \]
构造函数
\[\begin{align*} g(x) &= x - x^3, x \in \left[ 0, \frac{\sqrt{2}}{2} \right]\\ \end{align*} \] 其导函数 \(g'(x) = 1 - 3x^2\),则在 \(\left[0, \frac{\sqrt{3}}{3}\right]\) 上 \(g(x)\) 递增,在 \(\left[ \frac{\sqrt{3}}{3}, \frac{\sqrt{2}}{2} \right]\) 上 \(g(x)\) 递减。
即 \(r_1 + r_2\) 在 \(\cos \alpha = \frac{\sqrt{3}}{3}\) 时取得最小值 \(\frac{3 \sqrt{3}}{2}\)。由于 \(\sin \alpha \neq \cos \alpha\),则等号取不到。
- \(\alpha \in \left[ 0, \frac{\pi}{4} \right)\) 时 \(r_2 = \frac{1 - r_1 \sin^2 \alpha \cos \alpha}{\cos^2 \alpha \sin \alpha}\),则\[\begin{align*} r_1 + r_2 &= \frac{1}{\cos^2 \alpha \sin \alpha} + \frac{\cos \alpha - \sin \alpha}{\cos \alpha} r_1\\ &= \frac{1}{\cos^2 \alpha \sin \alpha} = \frac{1}{\sin \alpha - \sin^3 \alpha} \end{align*} \]同理,\(r_1 + r_2\) 在 \(\sin \alpha = \frac{\sqrt{3}}{3}\) 时取得最小值 \(\frac{3\sqrt{3}}{2}\)。由于 \(\cos \alpha \neq \sin \alpha\),则等号取不到。
综上所述,\(C_{ABCD} = 2 \left(r_1 + r_2\right) > 3 \sqrt{3}\)。
标签:cos,right,frac,解析几何,探究,alpha,平面,sin,left From: https://www.cnblogs.com/escap1st/p/17977305