JEE Main & Advanced
Mathematics
Straight Line
Question Bank
Self Evaluation Test - Straight Lines & Pair Striaght Lines
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 \[S=2\alpha \beta .\]