Question

已知椭圆C:x^2/a^2+y^2/b^2=1(a>b>0),过椭圆C右焦点F的直线y=x-1与椭圆C交于M.N两点，若x轴上存在点P使得

|PM|=|PN|,则|MN|/|PF|=?
A.2a B.2b C.2/a D.2/b

Answer 1

直线y=x-1过椭圆C右焦点F，所以F(1,0), c=1
椭圆C：x²/a²+y²/b²=1与直线y=x-1联立得
(a^2+b^2)x^2-2a^2x+a^2-a^2b^2=0
x1+x2=2a^2/(a^2+b^2),x1*x2=(a^2-a^2b^2)/(a^2+b^2)
|MN|=√(1+k^2)|x1-x2|
=√2*√[(x1+x2)^2-4x1x2]
=[2√2ab/(a^2+b^2)]*√(a^2+b^2-1) (a^2=b^2+c^2=b^2+1)
=[2√2ab/(a^2+b^2)]*√(2b^2)
=4ab^2/(a^2+b^2)
PM=PN,取MN中点Q( (x1+x2)/2, (y1+y2)/2 )
y1+y2=x1-1+x2-1=2a^2/(a^2+b^2) -2=-2b^2/(a^2+b^2)
Q(a^2/(a^2+b^2), -b^2/(a^2+b^2))
设P点坐标为(x0,0), PQ垂直MN, kPQ=-1
b^2/(a^2+b^2) / [x0- a^2/(a^2+b^2)] =-1
整理得x0=(a^2-b^2)/(a^2+b^2)=c^2/(a^2+b^2)=1/(a^2+b^2)
PF=1-1/(a^2+b^2)=(a^2+b^2-1)/(a^2+b^2)=2b^2/(a^2+b^2)
MN/PF=[4ab^2/(a^2+b^2)]/[2b^2/(a^2+b^2)]=2a
选A