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This repository was archived by the owner on Jul 21, 2026. It is now read-only.
https://hanatomizu.github.io/blog/post/%E5%85%B3%E4%BA%8E-%E8%A7%A3%E6%9E%90%E5%87%A0%E4%BD%95%E5%88%9D%E6%AD%A5-%E6%A4%AD%E5%9C%86-%E7%9A%84%E4%B8%80%E4%BA%9B%E5%A5%87%E6%8A%80%E6%B7%AB%E5%B7%A7/
本文章处于未完工状态(WIP)$a$ 为半长轴,$b$ 为半短轴,$c$ 为半焦距,满足 $a^2 = b^2 + c^2$ 椭圆方程为:$\displaystyle\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$A(x_1, y_1)$ , $B(x_2, y_2)$ 两点
$$\displaystyle |AB| = \sqrt{1+k^2}|x_2 - x_1| = \sqrt{1 + \frac{1}{k^2}}|y_2 - y_1|$$
$$\displaystyle |x_2 - x_1| = \frac{\sqrt{ -4(-a^2b^2)(b^2+a^2k^2 - m^2) }}{|b^2 + a^2k^2|}$$ $l_{AB}$ , 联立方程 具体过程 联立椭圆方程和斜线方程$a^2b^2$ 得到$y$ 带入
$$\displaystyle b^2x^2 + a^2(kx+m)^2 = a^2b^2$$
规定 椭圆中
斜线方程为:$f(x) = kx + m$
斜率的公式为:$\displaystyle k = \frac{y - y_1}{x - x_1}$
斜线与椭圆交于
结论 证明先放着
解题基本思路 通过斜线
$$\displaystyle \begin{cases} \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \newline y = kx + m \end{cases}$$ 椭圆方程同乘
$$\displaystyle \begin{cases} b^2x^2 + a^2y^2 = a^2b^2 \newline y = kx + m \end{cases}$$ 将