Question

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

• A. $\alpha \beta$
• B. $2\alpha \beta$
• C. $4\alpha \beta$
• D. None of these

Answer 1

The correct option is B $2\alpha \beta$

The equation of the given line

$\frac{x}{a}+\frac{y}{b}=1$ ...(1)

This line cuts x-axis at $A$ $(a,0)$ and $B$ $(0,b)$ respectively .

Since area of $△OAB=S$ (Given)

$\therefore \left|\frac{1}{2}ab\right|=S$

$ab=2S(\because ab>0)$ ...(2)

Since the line (1) passes through the point $P$ $(\alpha ,\beta )$

$\therefore \frac{\alpha }{a}+\frac{\beta }{b}=1\Rightarrow \frac{\alpha }{a}+\frac{a\beta }{2S}=1$ [Using(2)]

$\Rightarrow a^2\beta -2aS+2aS=0.$

Since a is real,

$\therefore 4S^2-8\alpha \beta S\ge 0$ $\Rightarrow 4S^2\ge 8\alpha \beta S\Rightarrow S\ge 2\alpha \beta$ $(\because S=\frac{1}{2}ab>0$ as $ab>)$

Hence, the least value of $S=2\alpha \beta .$