Let the line xa+yb=1 where ab>0, passes through fixed point P(α,β), where αβ>0. If the area formed by the line and the coordinate axes is S, then the least value of S can be expressed as λab where λ is
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Solution
Given equation of the line is xa+yb=1
So, the area of △OAB S=ab2(∵ab>0)
As the given line passes through P(α,β), αa+βb=1⇒bα+aβ=ab⇒(ab)α+a2β=a(ab)⇒βa2−2Sa+2Sα=0(∵ab=2S)
This is a quadratic equation in a. As a∈R, so D≥0⇒4S2−8αβS≥0⇒4S(S−2αβ)≥0
As S is area, so it cannot be negative,
therefore S≥2αβ