Question

The minimum area of the triangle formed by the variable line $3 \, \cos\,\theta \cdot x + 4 \,\sin \,\theta \cdot y = 12$ and the co-ordinate axes is

• A. $144$
• B. $\frac {25}{2}$
• C. $\frac {49}{4}$
• D. $12$

Answer 1

Answer: (D)

$12$

Answer 2

Given equation of line is
$x \cdot 3 \cos \theta+4 \,\sin \theta\, y=12$
$\Rightarrow \frac{x}{(4 / \cos \theta)}+\frac{y}{(3 / \sin \theta)}=1\,\,\,\,\,\,..(i)$
It intereset the coordinate axes at $A\left(\frac{4}{\cos \theta}, 0\right)$ and
$B\left(0, \frac{3}{\sin \theta}\right)$
$\therefore$ Area of $\Delta O A B$
$\Delta =\frac{1}{2} \times \frac{4}{\cos \theta} \times \frac{3}{\sin \theta}$
$=\frac{12}{\sin 2 \,\theta}\,\,\,\,\,\,\,\,..(ii)$
Now, for area to be minimum,
$\sin 2\, \theta$ should be maximum i.e.,
$\sin 2 \,\theta=1$
$\sin 2 \,\theta \mid \leq 1)\,\,\,\,\,\,\,\,\,(\because|\sin 2 \theta| \leq 1)$
$\therefore$ Minimum area
$\Delta_{\min }=\frac{12}{1}=12$