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)

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|求L的斜率

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本回答被提问者采纳

Answer 2

晕死