Question

椭圆方程为x^2/4y^2/2=1过右焦点F2的直线L交椭圆于AB两点若y轴上一点M(0,1/3)满足|MA|=|MB|

Answer 1

c^2=4-2=2

F2(√2,0)

AB:y=k(x-√2)

x^2/4+y^2/2=1

x^2+2[k(x-√2)]^2=4

(1+2k^2)x^2-4√2k^2*x+4k^2-4=0

AB的中点P

xP=(xA+xB)/2=2√2k^2/(1+2k^2)

yP=(yA+yB)/2=-√2k/(1+2k^2)

k(AB)*k(PM)=-1

k*(yP-1/3)/xP=-1

k(yP-1/3)=-xP

k*[-√2k/(1+2k^2)-1/3]=-2√2k^2/(1+2k^2)

2k^2-3√2k+1=0

k=(3√2±√10)/4