Question

已知椭圆x2/a2+y2/b2=1的离心率为根号2/2，短轴的一个端点为M(0,1）,直线l；y=

kx-1/3与椭圆相交于不同的两点A.B.若AB=4根号26/9，求k的值。求证；不论k取何值，以AB为直径的圆恒过点M

Answer 1

e^2=c^2/a^2=(a^2-b^2)/a^2=1-b^2/a^2=1/2

故有b^2/a^2=1/2
故有x^2/2b^2+y^2/b^2=1
A坐标代入得到4/2b^2+1/b^2=1,得到b^2=3
故椭圆方程是x^2/6+y^2/3=1
设经过B(3,0)的直线方程是x=my+3.
代入到椭圆中有m^2y^2+6my+9+2y^2-6=0
(m^2+2)y^2+6my+3=0
y1+y2=-6m/(m^2+2),y1y2=3/(m^2+2)
x1+x2=m(-6m/(m^2+2))+6=(-6m^2+6m^2+12)/(m^2+2)=12/(m^2+2)
x1x2=m^2y1y2+3m(y1+y2)+9=3m^2/(m^2+2)-18m^2/(m^2+2)+9=(3m^2-18m^2+9m^2+18)/(m^2+2)=(-6m^2+18)/(m^2+2)
k1=(y1-1)/(x1-2),k2=(y2-1)/(x2-2)
k1+k2=(y1-1)/(x1-2)+(y2-1)/(x2-2)
=[(y1-1)(x2-2)+(y2-1)*(x1-2)]/(x1-2)*(x2-2)
=[y1x2-2y1-x2+2+y2x1-2y2-x1+2]/[x1x2-2(x1+x2)+4]
=[y1*(my2+3)-2(y2+y1)-(x1+x2)+y2(my1+3)+4]/[(-6m^2+18)/(m^2+2)-24/(m^2+2)+4]
=[2m*3/(m^2+2)-6m/(m^2+2)-12/(m^2+2)+4]/[(-6m^2+18-24+4m^2+8)/(m^2+2)]
=[-12+4m^2+8]/[+2-2m^2]
=[-4(1-m^2)]/[2(1-m^2)]
=-2(为定值)