由题意,显然c=1,a=2 ,椭圆方程 x^2/4+y^2/3=1
设M(x1,y1),N(x2,y2),中点K(x0,y0) ; 直线MN :my+1 =x 代入椭圆方程 （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)
中垂线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 (因为不管m取什么值都与椭圆有2个交点）
4/m+3m ∈(-∞,-4√3 ]∪[4√3,+∞)
所以 y∈ [ -√3/12,√3/12]