一条线段AB长2a,两端点A和B分别在x轴和y轴上滑动,求线段AB中点M的轨迹方程.请简要写出思路及解答过程
人气:419 ℃ 时间:2020-04-14 13:45:56
解答
由题知,a>0
设M坐标(x,y)
则,A(2x,0) B(0,2y)
线段AB的长度为2a
可得方程:(2x)²+(2y)²=(2a)²
化简得 X^²+y²=a²
推荐
猜你喜欢
- what are you doing this weekend?选什么?
- 若sinx≥根号3/2,求x范围
- Looks up at the man that she turned
- 设数列{an{bn}{cn},已知a1=4,b1=3,c1=5,a(n+1)=an,b(n+1)=(an+cn)/2,c(n+1)=(an+bn)/2.求数列{cn-bn}的通项公式(2)求证:对任意n属于N*,bn+cn为定值
- it's tall and strong,but still it has no flowers.
- 甲乙两人从AB两地相向而行,相遇时,甲所行路程为乙的2倍多1.5千米,乙所行路程为甲路程的5/2,两地相距?
- 己知直线ax+4y-2=0与直线2x-5y+b=0互相垂直于点(1,c),求a,b,c,的值
- What will he do if he spend all the money?改错