已知标准方程为x2/a2+y2/b2=1(a>b>0)的椭圆C的左右焦点分别为F1,F2,上顶点为A,过点A作与AF2垂直的直线交x轴负半轴于点Q,2向量（F1F2）+向量（F2Q）=向量（0）,且过

题目

已知标准方程为x2/a2+y2/b2=1(a>b>0)的椭圆C的左右焦点分别为F1,F2,上顶点为A,过点A作与AF2垂直的直线交x轴负半轴于点Q,2向量（F1F2）+向量（F2Q）=向量（0）,且过A,Q,F2三点的圆恰好与直线l：x-根号（3）y-3=0相切.
（1）求椭圆C的离心率及椭圆C的方程
（2）已知点P（1,t）（t>0）是椭圆C上的定点,M,N是椭圆C上的两个动点,如果直线PM的斜率kPM与直线PN的斜率kPN满足kPM+kPN=0,试探究直线MN的斜率是否是定值?若是定值,求出这个定值,若不是,说明理由.

答案

(1)
F1 (-c,0),F2(c,0),A(0,b)
设Q(q,0)
向量AQ = (q,-b),向量AF2 = (c,-b)
AQ与AF2垂直,向量AQ•向量AF2 = cq + b² = 0,q = -b²/c
Q(-b²/c,0)
向量F1F2 = (2c,0),向量F2Q = (-b²/c - c,0)
2向量F1F2 +向量F2Q= (4c -b²/c - c,0) = 向量0
4c -b²/c - c = 0
b² = 3c²,b = √3c
a² = b² + c² = 4c²
a = 2c
e = c/a = 1/2
Q(-3c,0),F2(c,0),A(0,√3c)
F2Q的中点为M(-c,0)
QM = 2c = MF2
MA = √[(-c - 0)² + (0 - √3c)²] = 2c = QM
M为圆心,半径为2c
M与直线x-√3y-3=0的距离为r = 2c = |-c -0 - 3|/√(1 + 3) = (c+3)/2
c = 1
a = 2,b = √3
椭圆C的方程:x²/4 + y²/3 = 1
(2)
1/4 + t²/3 = 1
t = 3/2 (舍去-3/2)
设PM的斜率为k,PN的斜率为-k
PM的方程:y - 3/2 = k(x - 1)
与椭圆C的方程联立:(4k² + 3)x² + 4k(3 - 2k)x + (3 - 2k)² - 12 = 0
一个解为x₁ =1 (点P),x₁ + x2 = 4k(2k - 3)/(4k² + 3)
x₂ = 4k(2k - 3)/(4k² + 3) - x₁ = 4k(2k - 3)/(4k² + 3) - 1 = (4k² - 12k - 3)/(4k² + 3)
y₂ = k(x₂ - 1) + 3/2
PN的方程:y - 3/2 = -k(x - 1)
类似地可得N的坐标(x₃,y₃)
x₃ = (4k² + 12k - 3)/(4k² + 3)
y₃ = -k(x₃ - 1) + 3/2
MN的斜率K = (y₃ - y₂)/(x₃ - x₂)
x₃ + x₂ = (4k² - 12k - 3)/(4k² + 3) + (4k² + 12k - 3)/(4k² + 3) = (8k² - 6)/(4k² + 3)
y₃ - y₂ = -k(x₃ - 1) + 3/2 - k(x₂ - 1) - 3/2
= -k(x₃ + x₂) + 2k
= k [2 - (x₃ + x₂)]
= k[2 - (8k² - 6)/(4k² + 3)]
= 12k/(4k² + 3)
x₃ - x₂ = (4k² + 12k - 3)/(4k² + 3) - (4k² - 12k - 3)/(4k² + 3)
= 24k/(4k² + 3)
K = [12k/(4k² + 3)]/[24k/(4k² + 3)]
= 1/2
y₂ =