由題意,顯然c=1,a=2 ,橢圓方程 x^2/4+y^2/3=1
設(shè)M(x1,y1),N(x2,y2),中點K(x0,y0) ; 直線MN :my+1 =x 代入橢圓方程 （斜率1/m)
3(my+1)^2 +4x^2-12=0
(4+3m^2)y^2 +6my-9 = 0
所以 y0 = (y2+y1)/2 = -3m/(4+3m^2) ; x0 = my0+1 = 4/(4+3m^2)
中垂線KP:-m(x-4/(4+3m^2) = (y+3m/(4+3m^2)
x=0 ; y= m/(4+3m^2) = 1/(4/m+3m) m∈R (因為不管m取什么值都與橢圓有2個交點）
4/m+3m ∈(-∞,-4√3 ]∪[4√3,+∞)
所以 y∈ [ -√3/12,√3/12]