已知上x²/25+y²/9=1不同三点A(x1,y1)B(4,9/5)C(x2,y2)与F(4,0)的距离成等差数列,证明X1+X2=8；若线段AC的中垂线交X轴于点T,求直线BT的斜率
三点A(x1,y1)B(4,9/5)C(x2,y2)与F(4,0)的距离成等差数列
可知AF+CF=18/5
根据椭圆性质到焦点距离与到准线的距离比为离心率e
显然F点为椭圆的焦点.
则AF=e(a²/c-x1),CF=e(a²/c-x2)
则AF+CF=5-4x1/5+5-4x2/5=18/5
x1+x2=8
x1²/25+y1²/9-x2²/25-y2²/9=0
(x1+x2)/25+(y1+y2)(y1-y2/9(x1-x2)
则k=-9（x1+x2)/25(y1+y2)
AC垂直平分线为：y-(y1+y2)/2=25(y1+y2)(x-4)/72
令y=0,则x=4-36/25=364/25
则kbt=9/5/(4-364/25)=5/4