c/a=√3/2,∴c^2/a^2=3/4,
∴b^2/a^2=1/4,a=2b,
设椭圆方程为x^2/(4b^2)+y^2/b^2=1,b>0,椭圆上任意一点M为(2bcosu,bsinu),
PM^2=4b^2(cosu)^2+(3/2-bsinu)^2
=4b^2-4b^2(sinu)^2+9/4-3bsinu+b^2(sinu)^2
=-3b^2(sinu)^2-3bsinu+4b^2+9/4
=-3b^2[sinu-1/(2b)]^2+4b^2+3,
1/(2b)=1/2时4b^2+3=7,b=1,sinu=1/2,cosu=土√3/2,
这时,椭圆方程是x^2/4+y^2=1,所求点为（土√3,1/2）；
1/（2b)>1,即