Question

Minimum area of the triangle which a member of this family with negative gradient can make with the positive semi axes, is?

• A. $8$
• B. $6$
• C. $4$
• D. $2$

Answer 1

The correct option is C $4$

According to the problem

$\left(4a+3\right)-\left(a+1\right)y-\left(2a+1\right)=0$ where $a\in R$

$y=\frac{\left(4a+3\right)}{\left(a+1\right)}\times -\frac{\left(2a+1\right)}{\left(a+1\right)}\dots \dots \left(i\right)$

condition: The line has negative gradient

$\therefore \frac{\left(4a+3\right)}{\left(a+1\right)}<0$

$\therefore a\in \left(-1,-\frac{3}{4}\right)$

The $x$ intercept of given line : at $y=0$

$x=\frac{\left(2a+1\right)}{\left(4a+3\right)}$

The $y$ intercept of given line : at $x=0$

$x=-\frac{\left(2a+1\right)}{\left(a+1\right)}$

Area of triangle formed is

$\Delta =\frac{1}{2}\times$($x$-intercept)$\times$($y$-intercept)

$\Rightarrow \Delta =\frac{1}{2}\times \frac{\left(2a+1\right)}{\left(4a+3\right)}\times -\frac{\left(2a+1\right)}{\left(a+1\right)}$

$\Rightarrow \Delta =-\frac{1}{2}\times \frac{{\left(2a+1\right)}^2}{\left(4a^2+7a+3\right)}$

for min $\Delta ,\frac{d\Delta }{da}$ must be zero.

$\Delta =-\frac{1}{2}\times \frac{{\left(2a+1\right)}^2}{\left(4a^2+7a+3\right)}$

differentiating w.r.t $a$

$\frac{d\left(\Delta \right)}{da}=-\frac{1}{2}\times \frac{2{\left(2a+1\right)}^2\times \left(2.1\right)\times \left(4a^2+7a+3\right)-{\left(2a+1\right)}^2\times \left(8a+7\right)}{{\left(4a^2+7a+3\right)}^2}$

$\Rightarrow 0=4\left(2a+1\right)\left(4a^2+7a+3\right)-{\left(2a+1\right)}^2\times (8a+7)$

$\Rightarrow 0=\left(2a+1\right)\left\{\left(16a^2+28a+12\right)-\left(2a+1\right)(8a+7)\right\}$

$\Rightarrow 0=\left(2a+1\right)(16a^2+28a+12-16a^2-22a-7)$

$\Rightarrow 0=\left(2a+1\right)(6a+5)$

$\therefore$ either $a=-\frac{1}{2}$ or $a=-\frac{5}{6}$

But $a$ cannot be $-\frac{1}{2}$ as $a\in \left(-1,-\frac{3}{4}\right]$ for negative gradient.

$\therefore$ for $a=-\frac{5}{6},$ area of triangle formed is minimum.

$\therefore \Delta min=-\frac{1}{2}\times \frac{{\left(2a+1\right)}^2}{\left(4a^2+7a+3\right)}$ at $a=-\frac{5}{6}$

$=-\frac{1}{2}\times \frac{{\left(-\frac{5}{6}\times 2+1\right)}^2}{\left(4\times \frac{25}{36}-\frac{5}{6}\times 7+3\right)}$

$=-\frac{1}{2}\times \frac{{\left(-\frac{2}{3}\right)}^2}{\left(-\frac{2}{36}\right)}$

$=-\frac{1}{2}\times \frac{4}{9}\times \frac{36}{-2}$

$=4$

$\therefore$ Area of the required triangle is $4$.

Hence the correct option is $C$.