已知椭圆 x 2 2 +y 2 =1 ，直线 y=x+1 与椭圆交于 A 、 B 两点，求 |AB| 。 4 2 3 直线与椭圆的位置关系直线与椭圆韦达定理弦长公式全国乙卷 2023

3 4 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"/> 联立并消元：把 y = x + 1 代入椭圆方程得 2 x 2 + ( x + 1 ) 2 = 1 ，整理为 3 x 2 + 4 x = 0 。由韦达定理： x 1 + x 2 = − 3 4 ， x 1 x 2 = 0 。 ( x 1 − x 2 ) 2 = ( x 1 + x 2 ) 2 − 4 x 1 x 2 = 9 16 。弦长公式： ∣ A B ∣ = 1 + k 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"/> ∣ x 1 − x 2 ∣ = 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"/> × 3 4 = 3 4 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"/> 。

高考数学考点题库/椭圆/直线与椭圆的位置关系

2023 年全国乙卷·第20题

解答题★★★★☆ 较难改编题通法

已知椭圆

\frac{x ^{2}}{2} + y^{2} = 1

，直线

y = x + 1

与椭圆交于

A

、

B

两点，求

∣ A B ∣

。

答案

\frac{4 2}{3}

分步解析

1联立并消元：把 $y = x + 1$ 代入椭圆方程得 $\frac{x ^{2}}{2} + (x + 1)^{2} = 1$ ，整理为 $3 x^{2} + 4 x = 0$ 。
2由韦达定理： $x_{1} + x_{2} = - \frac{4}{3}$ ， $x_{1} x_{2} = 0$ 。
3 $(x_{1} - x_{2})^{2} = (x_{1} + x_{2})^{2} - 4 x_{1} x_{2} = \frac{16}{9}$ 。
4弦长公式： $∣ A B ∣ = 1 + k^{2} ∣ x_{1} - x_{2} ∣ = 2 \times \frac{4}{3} = \frac{4 2}{3}$ 。

点评弦长公式

∣ A B ∣ = 1 + k^{2} ∣ x_{1} - x_{2} ∣

中的

∣ x_{1} - x_{2} ∣

一律通过韦达定理算

(x_{1} + x_{2})^{2} - 4 x_{1} x_{2}

，不要去解方程求根。

#直线与椭圆#韦达定理#弦长公式

2023 年真题全国乙卷

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