已知椭圆的一个焦点与短轴两端点构成等边三角形，则其离心率为＿＿＿＿＿＿。 3 2 离心率离心率几何条件等边三角形北京卷 2022

2 3 <path d="M95,702 c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14 c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54 c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10 s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429 c69,-144,104.5,-217.7,106.5,-221 l0 -0 c5.3,-9.3,12,-14,20,-14 H400000v40H845.2724 s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7 c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z M834 80h400000v40h-400000z"/> 设焦点 F ( c , 0 ) ，短轴端点 B 1 ( 0 , b ) 、 B 2 ( 0 , − b ) 。由对称性 ∣ F B 1 ∣ = ∣ F B 2 ∣ = b 2 + c 2 <path d="M95,702 c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14 c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54 c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10 s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429 c69,-144,104.5,-217.7,106.5,-221 l0 -0 c5.3,-9.3,12,-14,20,-14 H400000v40H845.2724 s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7 c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z M834 80h400000v40h-400000z"/> = a ，故只需 ∣ B 1 B 2 ∣ = 2 b 也等于 a 。由等边条件 2 b = a ，即 b = 2 a 。代入 c 2 = a 2 − b 2 = a 2 − 4 a 2 = 4 3 a 2 ，得 e = a c = 2 3 <path d="M95,702 c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14 c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54 c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10 s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429 c69,-144,104.5,-217.7,106.5,-221 l0 -0 c5.3,-9.3,12,-14,20,-14 H400000v40H845.2724 s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7 c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z M834 80h400000v40h-400000z"/> 。

2022 北京卷｜椭圆·离心率｜天府数学实验室

已知椭圆的一个焦点与短轴两端点构成等边三角形，则其离心率为＿＿＿＿＿＿。

答案

\frac{3}{2}

分步解析

1设焦点 $F (c, 0)$ ，短轴端点 $B_{1} (0, b)$ 、 $B_{2} (0, - b)$ 。
2由对称性 $∣ F B_{1} ∣ = ∣ F B_{2} ∣ = b^{2} + c^{2} = a$ ，故只需 $∣ B_{1} B_{2} ∣ = 2 b$ 也等于 $a$ 。
3由等边条件 $2 b = a$ ，即 $b = \frac{a}{2}$ 。
4代入 $c^{2} = a^{2} - b^{2} = a^{2} - \frac{a ^{2}}{4} = \frac{3 a ^{2}}{4}$ ，得 $e = \frac{c}{a} = \frac{3}{2}$ 。

点评「焦点与短轴端点」的连线长恒为

a

，这一条结论能把很多几何条件迅速转化成

a

、

b

、

c

的方程。

#离心率#几何条件#等边三角形

2022 年真题北京卷

2022 年北京卷·第16题

分步解析

同考点推荐椭圆 · 离心率