Question

已知椭圆C的两焦点为F1(-1,0),F2(1,0),并且经过点(1,√2/2),直线l过点F2与

Answer 1

c=1,2a=√[2^2+1/2]+1/√2=3/√2+1/√2=2√2,a=√2,b=1,

∴椭圆方程是x^2/2+y^2=1,①

l:x=my+1,②

代入①,m^2y^2+2my+1+2y^2=2,

整理得(m^2+2)y^2+2my-1=0,

设交点A(x1,y1),B(x2,y2),则y1+y2=-2m/(m^2+2),y1y2=-1/(m^2+2),

由②,x1x2=(my1+1)(my2+1)=m^2*y1y2+m(y1+y2)+1,

∴向量OA*OB=x1x2+y1y2=(m^2+1)y1y2+m(y1+y2)+1

=(-m^2-1-2m^2)/(m^2+2)+1

=5/(m^2+2)-2,

u=m^2+2的值域是[2,+∞),

5/u-2的值域是(-2,1/2],为所求.