设经点F(1,0)的直线l与椭圆x^2/2+y^2=1交于AB 求以OA OB 为邻边的平行四边形OABP的顶点P的轨迹?

解;设直线为y=k(x-1),联立方程组有:(1+2k*k)x*x-4k*kx+2k*k-2=0则：x1+x2=4k*k/(1+2k*k)以OA OB 为邻边的平行四边形OABP的顶点P,存在向量OA+向量OB=向量OP设A(x1,y1),B(x2,y2)；P(x,y) 则x=x1+x2=4k*k/(1+2k*k) (1)y...