Through the point P(α,β) , where αβ>0 , the straight line xa+yb=1 is drawn so as to form with axes a triangle of  area S . If ab>0, then least value of S is

Area of ΔOAB=S=12ab----(1) Equation of AB is xa+yb=1Putting α,β, we get αa+βb=1from (1) b=2Sa⇒αa+aβ2S=1------(2)⇒a2β−2aS+2αS=0∴ a∈R⇒ D≥04S2−8αβS≥0S≥2αβ Least value of S=2αβ