1、当P点在右顶点时二向量积有最大值,
c=√（4-3）=1,
OP•FP=|a+c|*|a|*cos0°=|（2+1）|*2=6.
2、c^2=a^2-b^2=1,c=1,
直线方程为：y=2(x-1),2x-y-2=0,
原点至直线距离d=|0-0-2|/√（4+1）=2/√5,
x^2/5+(2x-2)^2/4=1,
3x^2-5x=0,
x1=0,y1=-2,
x2=5/3,y2=4/3,
|AB|=√[(x1-x2)^2+(y1-y2)^2]=5√5/3,
S△AOB=|AB|*d/2=5/3.
以BA为X轴,BA中点的垂线为Y轴,建立直角坐标系,
则B（-c,0）,A(c,0),
|BA|=2c,根据椭圆定义,|CA|+|CB|=2a,
|CA|/|CB|+1=2a/|CB|,
tanB+1=2a/|CB|,
3/4+1=2a/|CB|,
e=c/a,a=c/e,
secB=√(1+(tanB)^2)=5/4,
cosB=4/5,
7/4=2c/(e*|CB|=2/(e*CB/c)=1/(e*|CB|/2c)=1/(e*cosB)=1/(4e/5),
5/(4e)=7/4,
e=5/7,
离心率e=5/7.
4、 设椭圆方程为：x^2/a^2+y^2/b^2=1,
左右焦点坐标F1（-c,0）,F2(c,0),
P(x0,y0),
(x0+c)^2+y0^2=48,(1)
(x0-c)^2+y0^2=12,(2)
(1)-(2)式,
4cx0=36,
x0=9/c,
根据三角形角平分线比例线段的性质,
|PF1|/|PF2|=|F1Q|/|QF2|,
|F1Q|=1+c,|QF2|=c-1,
4√3/（2√3）=（1+c）/(c-1),
c=3,
x0=9/3=3,
y0=±2√3,
b^2=a^2-c^2=a^2-9,
椭圆方程为：x^2/a^2+y^2/(a^2-9)=1,
将P点坐标代入椭圆方程求出a,
a^4-30a^2+81=0,
(a^2-27)(a^2-3)=0,
a^2=27,a^2=3,
因c=3,故舍去a^2=3,
a^2=27,
则椭圆方程为：x^2/27+y^2/18=1.
5、设椭圆方程为x^2/a^2+y^2/b^2=1,(a>b>0),暂设焦点在X轴,
e=c/a=√3/2,c=√3a/2,b^2=a^2-c^2=a^2/4,
椭圆方程为:x^2/a^2+y^2/(a^2/4)=1,
y=-x/2-4,代入椭圆方程,
2x^2+16x+64-a^2=0,
根据韦达定理,
x1+x2=-8,x1*x2=32-a^2/2,
根据弦长公式,
|PQ|=√（1+k^2）(x1-x2)^2
=√(1+1/4)[(x1+x2)^2-4x1*x2]
=(1/2)√[5*(64-128+2a^2)]
=(1/2)√(10a^2-320)
=√10,
a=6,
b=3,
椭圆方程为：x^2/36+y^2/9=1,
若焦点在Y轴,一样可做.
6、设A（x1,y1）,B(x2,y2),AB直线斜率k,
x1^2/16+y1^2/4=1,(1)
x2^2/16+y2^2/4=1,(2)
(1)-(2)式,
(x1^2-x2^2)/16+(y1^2-y2^2)/4=0,
1/4+[(y1-y2)/(x1-x2)]/{[(y1+y2)/2]/[(x1+x2)/2]}=0,
k=( y1-y2)/(x1-x2),
(y1+y2)/2和(x1+x2)/2为M点纵、横坐标,
1/4+k*(1/2)=0,
k=-1/2,
则直线方程为：(y-1)/(x-2)=-1/2,
即x+2y-4=0.
故存在这样的直线.
