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Q.

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

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a

αβ

b

2αβ

c

3αβ

d

none

answer is B.

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Detailed Solution

OA=OQ+QA=α+βcotθOB=OR+RB=β+αtanθNow, the area of ΔOAB is 12α+βcotθβ+αtanθ=122αβ+α2tanθ+β2cotθ=122αβ+αtanθ−βcotθ+2αβ≥122αβ+2αβ=2αβLeast value of S=2αβ
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