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 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

Answer 1

Answer: (B)

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