Question

The area of the region bounded by the curves $|x+y|\le 2,|x-y|\le 2$ and $2x^2+6y^2\ge 3$ is

• A. $(8+\frac{\sqrt{3}}{2}\pi )sq.units$
  
• B. $(8-\frac{\sqrt{3}}{2}\pi )sq.units$
  
• C. $(4-\frac{3\sqrt{3}}{2}\pi )sq.units$
  
• D. $(8-\frac{3\sqrt{3}}{2}\pi )sq.units$

Answer 1

The correct option is B $(8-\frac{\sqrt{3}}{2}\pi )sq.units$




$2x^2+6y^2\ge 3$
area of ellipse = $\pi \times \frac{\sqrt{3}}{\sqrt{2}}\times \frac{1}{\sqrt{2}}=\frac{\pi \sqrt{3}}{2}$ .......(1)
$|x+y|\le 2\Rightarrow -2\le (x+y)\le 2$ ......(2)
$|x-y|\le 2\Rightarrow -2\le (x-y)\le 2$ .......(3)
Required area = $(8-\frac{\pi \sqrt{3}}{2})$ sq. units