一条变动的直线L与椭圆X＾2／4＋Y＾2／2＝1交于P,Q两点,M是L上的动点,满足关系｜MP｜×｜MQ｜＝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

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

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[...