设所求方程是：x^2/a^2+y^2/b^2=1,(a>b>0),根据题意：b=1,c/a=√3/2,因为b^2=a^2-c^2,所以：1=a^2-c^2,即：a^2=1+c^2,由c/a=√3/2得：c^2=3a^2/4,所以：a^2=1+3a^2/4,解得：a^2=4,故：所求方程是：x^2/4+y^2=1...