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 $a b > 0$, 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 $\triangle O A B=S=\frac{1}{2} a b$ ..... (1)
Equation of $A B$ is $\frac{x}{a}+\frac{y}{b}=1$
Putting $(\alpha, \beta)$, we get
$ \frac{\alpha}{a}+\frac{\beta}{b}=1$
$\Rightarrow \frac{\alpha}{a}+\frac{a \beta}{2 S}=1$[Using (1)]
$\Rightarrow a^{2} \beta-2 a S+2 \alpha S=0$
$\therefore a \in R $
$\Rightarrow D \geq 0 $
$ 4 S^{2}-8 \alpha \beta S \geq 0$
$ S \geq 2 \alpha \beta$
Least value of $S=2 \alpha \beta$