Question

设f1,f2分别为椭圆c的左右焦点,过点f2的直线l与椭圆相交于a b两点60°

Answer 1

设椭圆C:x^2/a^2+y^2/b^2=1(a>b>0)的焦点F1(-c,0),F2(c,0),

由AF2=2F2B（改题了）得

B(c+h,√3h),A(c-2h,-2√3h)在椭圆上,

∴(c+h)^2/a^2+3h^2/b^2=1,①

(c-2h)^2/a^2+12h^2/b^2=1,②

②-①,(3h^2-6ch)/a^2+9h^2/b^2=0,h≠0,

∴(h-2c)/a^2+3h/b^2=0,(3a^2+b^2)h=2b^2c,h=2b^2c/(3a^2+b^2),

代入①,9c^2(a^2+b^2)^2+12a^2b^2c^2=a^2(3a^2+b^2)^2,c^2=a^2-b^2,

∴(a^2-b^2)(9a^4+30a^2b^2+9b^4)=a^2(9a^4+6a^2b^2+b^4),

化简得15a^4-22a^2b^2-9b^4=0,

∴a^2=9b^2/5,c=2b/√5,

F1(-2b/√5,0）到直线l:y=√3(x-2b/√5)的距离=2b√(3/5)=2√3,

∴b=√5,a^2=9,

∴椭圆C的方程是x^2/9+y^2/5=1.