question_answer

Through the point \[P(\alpha ,\beta )\], where \[\alpha \beta >0.\] the straight line \[\frac{x}{a}+\frac{y}{b}=1\] is drawn so as the form with axes a triangle of area S. if \[ab>0,\] then least value of S is

A) \[\alpha \beta \]

B) \[2\alpha \beta \]

C) \[3\alpha \beta \]

D) None of these

Correct Answer: B

Solution :

[b] Area of \[\Delta OAB=S=\frac{1}{2}ab\] Equation of AB is \[\frac{x}{a}+\frac{y}{b}=1\] Putting \[(\alpha ,\beta ),\] we get \[\frac{\alpha }{a}+\frac{\beta }{b}=1\] \[\Rightarrow \frac{\alpha }{a}+\frac{\alpha \beta }{2S}=1\]                                 [using (i)] \[\Rightarrow {{a}^{2}}\beta -2aS+2aS=0\]\[\therefore a\in R\Rightarrow D\ge 0\] \[4{{S}^{2}}-8\alpha \beta S\ge 0\] \[\Rightarrow S\ge 2\alpha \beta .\] Least value of \[S=2\alpha \beta .\]