椭圆C:x^2/a^2+y^2/b^2=1(a>b>0),
离心率为根号3/2,
过点M(2,1),
则,{ a^2=b^2+c^2
c/a=根号3/2
4/a^2+1/b^2=1
解得a^2=8,b^2=2
所以,椭圆方程为x^2/8+y^2/2=1
第二题意思不清,A,B两点是不是直线l与椭圆的交点?
直线l∥OM,OM斜率为1/2
设直线l的方程为y=1/2x+p,
代入椭圆方程,得x^2+2px+2p^2-4=0
设A(x1,y1),B(x2,y2)
则,x1+x2=-2p,x1*x2=2p^2-4,
直线MA,MB的斜率分别为k1,k2.
则,k1=(y1-1)/(x1-2),k2=(y2-1)/(x2-2)
k1+k2
=(y1-1)/(x1-2)+(y2-1)/(x2-2)
=(1/2x1+p-1)/(x1-2)+(1/2x2+p-1)/(x2-2)
=[x1*x2+(x1+x2)*(p-2)-4p+4]/[(x1-2)(x2-2)]
=[2p^2-4+(-2p)*(p-2)-4p+4]/[(x1-2)(x2-2)]
=0