Through the point P( α α , β β ), where α α β β > 0, the straight line 
is drawn so as to form with coordinate axes a triangle of area s. If ab> 0, then least value of s is
Text Solution
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Given P ≡ ≡ ( α α , β β ).
Given line is 
.. . . . . (1)
If line (1) cuts x and y axes at A and B respectively, then
A ≡ ≡ (a, 0) and B ≡ ≡ (0, b). Also the area of Δ Δ OAB = s
i.e. (1/2)ab = s ⇒ ⇒ ab = 2s.
Since line (1) passes through P( α α , β β ),
⇒ ⇒ 
⇒ ⇒ a 2 β β – 2as + 2 α α s = 0.
Since a is real, 4s 2 – 8 α α β β s ≥ ≥ 0 ⇒ ⇒ s ≥ ≥ 2 α α β β .
Hence the least value of s = 2 α α β β .
Hence is the correct answer.
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