Question

在平行四边形ABCD中,AB=4 AD=3 E是AB上一动点 AE=x DE的延长线交CB延长线于F 设CF=y

Answer 1

y=BC+FB = 3 + FB

FB : AD = EB :AE

FB = EB*AD/AE = 3(4-x)/x

y = 3 + 12/x - 3 = 12/x , 0