Question

椭圆C:x^2/a^2+y^2/b^2=1(a>b>0)的离心率e=√3/2,且过(2,0)点.

Answer 1

因为是x^2/a^2+y^2/b^2=1且过（2,0）点所以a=2

e=√3/2所以c=√3所以b=1

所以方程为x^2/4+y^2=1

设A(x1,y1) B(x2,y2)

x^2/4+y^2=1

y=x+m 联立 得

x^2/4+x^2+2mx+m^2-1=0

5x^2+8mx+4(m^2-1)=0

△OAB直角三角形

1.OA⊥OB

则x1x2+y1y2=0

x1x2=4(m^2-1)/5

y1y2=x1x2+m(x1+x2)+m^2

x1x2+y1y2=8(m^2-1)/5-m(8m)/5+m^2=0

-8/5+m^2=0

m=±2√10/5

2.AB⊥OA

kAB=1 kOA=-1

直线OA经过原点 方程为 y=-x

y=-x

x^2/4+y^2=1

x^2/4+x^2=1

5x^2=4 x=±2√5/5

x=2√5/5 y=-2√5/5 m=-4√5/5

x=-2√5/5 y=2√5/5 m=4√5/5