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Question

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=1bα+aβ=ab(ab)α+a2β=a(ab)βa22Sa+2Sα=0 (ab=2S)
This is a quadratic equation in a. As aR, so
D04S28αβS04S(S2αβ)0
As S is area, so it cannot be negative,
therefore S2αβ

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