过椭圆x2/a2+y2/b2=1(a>b>0)的一个焦点F作弦AB,若AF=d1,BF=d2,求1/d1+1/d2的值?

c/xA=BF/AB=a/AB
AB=a*xA/c=2a^3/(a^2+c^2)
d1=AF=AB-BF=2a^3/(a^2+c^2)-a=a(a^2-c^2)/(a^2+c^2)=ab^2/(a^2+c^2)
1/d1+1/d2=(a^2+c^2)/(ab^2)+1/a=(a^2+c^2+b^2)/(ab^2)=2a^2/(ab^2)=2a/b^2
答：1/d1+1/d2=2a/b^2