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$Through the point P (α, β), where α β > 0, the straight line \frac{x}{a} + \frac{y}{b} = 1 is drawn so as to form with axes a triangle of$
$area S . If a b > 0, then least value of S is$

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Correct option is B

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

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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

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Ready to Test Your Skills?

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$Through the point P (α, β), where α β > 0, the straight line \frac{x}{a} + \frac{y}{b} = 1 is drawn so as to form with axes a triangle of$
$area S . If a b > 0, then least value of S is$