Question

Let the line $\frac{x}{a}+\frac{y}{b}=1$ where $ab>0,$ passes through $P(\alpha ,\beta ),$ where $\alpha \beta >0.$ If the area formed by the line and the coordinate axes is $S,$ then the least value of $S$ is

• A. $\alpha \beta$
• B. $2\alpha \beta$
• C. $4\alpha \beta$
• D. $\frac{\alpha \beta }{2}$

Answer 1

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

Given equation of the line is $\frac{x}{a}+\frac{y}{b}=1$

So, the area of $△OAB$
$S=\frac{ab}{2}(\because ab>0)$
As the given line passes through $P(\alpha ,\beta )$,
$\frac{\alpha }{a}+\frac{\beta }{b}=1\Rightarrow b\alpha +a\beta =ab\Rightarrow \left(ab\right)\alpha +a^2\beta =a\left(ab\right)\Rightarrow \beta a^2-2Sa+2S\alpha =0(\because ab=2S)$
This is a quadratic equation in $a.$ As $a\in R$, so
$D\ge 0\Rightarrow 4S^2-8\alpha \beta S\ge 0\Rightarrow 4S(S-2\alpha \beta )\ge 0$
As $S$ is area, so it cannot be negative,
therefore $S\ge 2\alpha \beta$