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)

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 $\Delta OAB$
$\Delta =\frac{1}{2}\times \frac{4}{\cos \theta }\times \frac{3}{\sin \theta }$
$=\frac{12}{\sin 2\theta }..(ii)$
Now, for area to be minimum,
$\sin 2\theta$ should be maximum i.e.,
$\sin 2\theta =1$
$\sin 2\theta \mid \le 1)(\because |\sin 2\theta |\le 1)$
$\therefore$ Minimum area
${\Delta }_{\min }=\frac{12}{1}=12$