- 一条变动的直线L与椭圆X^2/4+Y^2/2=1交于P,Q两点,?
- 接上:MP的绝对值*MQ的绝对值=2,若直线L在变动过程中始终保持其斜率等于1,求动点M的轨迹方程,并说明曲线的形状。
- 直线L:y=x+k, P(x1,y1), Q(x2,y2), M(m,n)
==> y1=x1+k, y2=x2+k, k=n-m ...(1)
y=x+k 代入X^2/4+Y^2/2=1,得:
3x^2 +4kx +(2k^2 -4) =0 ...(2)
|MP|*|MQ|=2
|MP| =根号[(x1-m)^2+(y1-n)^2] =根号[2*(x1-m)^2]
|MQ| =根号[(x2-m)^2+(y2-n)^2] =根号[2*(x2-m)^2]
1 =|(x1-m)(x2-m)| =|x1x2 -(x1+x2)m +m^2| ...(3)
(1)(2)(3) ==>
m^2 +2*n^2 = 1,or 7
因此,动点M的轨迹为椭圆:
x^2 +2*y^2 =1, 及:x^2 +2*y^2 =7