已知椭圆 x 2 4 + y 2 3 =1 ，点 P 在椭圆上，求 OAP 面积的最大值，其中 A(2,0) ， O 为原点。 3 面积最值面积最值椭圆取值范围新课标I卷 2022

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"/> 以 O A 为底， ∣ O A ∣ = 2 ，则面积 S = 2 1 × 2 × ∣ y P ∣ = ∣ y P ∣ 。点 P 在椭圆上，纵坐标满足 ∣ y P ∣ ≤ b = 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"/> 。当 P 取短轴端点 ( 0 , ± 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"/> ) 时 ∣ y P ∣ 最大。故面积最大值为 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"/> 。

高考数学考点题库/圆锥曲线综合定点定值/面积最值

2022 年新课标I卷·第20题

解答题★★★★★ 压轴改编题巧法

已知椭圆

\frac{x ^{2}}{4} + \frac{y ^{2}}{3} = 1

，点

P

在椭圆上，求

△ O A P

面积的最大值，其中

A (2, 0)

，

O

为原点。

答案

3

分步解析

1以 $O A$ 为底， $∣ O A ∣ = 2$ ，则面积 $S = \frac{1}{2} \times 2 \times ∣ y_{P} ∣ = ∣ y_{P} ∣$ 。
2点 $P$ 在椭圆上，纵坐标满足 $∣ y_{P} ∣ \leq b = 3$ 。
3当 $P$ 取短轴端点 $(0, \pm 3)$ 时 $∣ y_{P} ∣$ 最大。
4故面积最大值为 $3$ 。

点评底边固定在坐标轴上时，面积只取决于另一点到该轴的距离，直接用椭圆的取值范围即可，不必设直线。

#面积最值#椭圆#取值范围

2022 年真题新课标I卷

同考点推荐圆锥曲线综合定点定值 · 面积最值

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2022 年新课标I卷·第20题

分步解析

同考点推荐圆锥曲线综合 定点定值 · 面积最值

同考点推荐圆锥曲线综合定点定值 · 面积最值