Question

已知椭圆x2 a2+y2 b2 1的焦距为4,且与椭圆x2+y2\2=1有相同的离心率，斜率为k的

Answer 1

（1）对椭圆
x^2+y^2/2=1,
a1=√2，b1=1，c1=1
对椭圆
x^2/a^2+y^2/b^2=1,
焦距2c=4
=>
c=2
有相同离心率，则
e=c1/a1=c/a=1/√2=2/a
=>
a=2√2
b^2=a^2-c^2=8-4=4
∴椭圆C的方程为：x^2/8+y^2/4=1
（2）设直线方程为
y=kx+1
代入椭圆C可得，x^2/8+(kx+1)^2/4=1
整理得
(1+2k^2)x^2+4kx-6=0
设交点A(x1,y1),
B(x2,y2)，则有
x1+x2=-4k/(1+2k^2),
x1x2=-6/(1+2k^2)
y1+y2=k(x1+x2)+2=2/(1+2k^2)
y1y2=k^2x1x2+k(x1+x2)+1=(1-8k^2)/(1+2k^2)
设AB中点为M，右焦点为F(2,0)，则有
M=M((x1+x2)/2,(y1+y2)/2)=M(-2k/(1+2k^2),1/(1+2k^2))
AB=√[(x1-x2)^2+(y1-y2)^2]
=√[(x1+x2)^2-4x1x2+(y1+y2)^2-4y1y2]
=√[(-4k/(1+2k^2))^2+4*6/(1+2k^2)+(2/(1+2k^2))^2-4(1-8k^2)/(1+2k^2)]
=√[(16k^2+4)/(1+2k^2)^2+4(5+8k^2)/(1+2k^2)]
=√4/(1+2k^2)^2*[(4k^2+1)+(5+8k^2)(1+2k^2)]
=2/(1+2k^2)*√[(16k^4+22k^2+6)]
FM=√[(2+2k/(1+2k^2))^2+(0-1/(1+2k^2))^2]
=√[((2+2k+4k^2)/(1+2k^2))^2+(1/(1+2k^2))^2]
=1/(1+2k^2)*√[(2+2k+4k^2)^2+1]
右焦点F在以AB为直径的圆内，则有
FM
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