> 数学 >
设线段AB的两个端点A、B分别在x轴、y轴上滑动,且|AB|=5且向量OM=3/5向量OA+2/5向量OB,求点M的轨迹方程
人气:395 ℃ 时间:2019-11-02 01:40:24
解答
∵A、B分别在x轴、y轴上设A(a,0),B(0,b),M(x,y)且|AB| = 5则√[(a-0)² + (0-b)²] = 5得a² + b² = 25 (1)又OM = (3/5)OA + (2/5)OB(x,y) = (3/5)(a,0) + (2/5)(0,b)(x,y) = (3a/5,2b/5)得x = 3a/...
推荐
猜你喜欢
© 2026 79432.Com All Rights Reserved.
电脑版|手机版