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 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

Answer: (B)

$2\alpha \beta$

The equation of the given line is

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

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

Area of $△AOB=S$ (given)

$\therefore \left|\frac{1}{2}ab\right|=S$ or $ab=2S(\because ab>0)...\left(i\right)$

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

$\frac{\alpha }{a}+\frac{\beta }{b}=1$

or $\frac{\alpha }{a}+\frac{a\beta }{2S}=1$ (Using eq. (i))

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

since $a$ is real

$\therefore 4S^2-8\alpha \beta S\ge 0$

$\Rightarrow S\ge 2\alpha \beta [\because S=\frac{1}{2}ab>0]$

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