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

The equation of the given line isxa+yb=1                                         (1)This line cuts x-axis and y-axis at A(a, 0) and B(0, b) respectively.Since area of ∆OAB = S (Given)∴12ab=S  or ab=2S ∵ab > 0      (2)Since the line (1) passes through the point P(α,β)∴αa+βb=1  or  αa+αβ2S=1     [Using (2)]or  α2β-2aS+2αS=0Since a is real, ∴4S2-8αβS≥0or  4S2≥8αβS  or  S≥2αβ    ∵S=12ab>0  as  ab > 0Hence the least value of S=2αβ