Question

The area of the region inside the parabola $5x^2-y=0$ but outside the parabola $2x^2-y+9=0$ is

• A. $12\sqrt{3}$ sq. units
• B. $6\sqrt{3}$ sq. units
• C. $8\sqrt{3}$ sq. units
• D. $4\sqrt{3}$ sq. units

Answer 1

The correct option is A $12\sqrt{3}$ sq. units

Given $5x^2-y=0$
$2x^2-y+9=0$
Eliminating $y$, we get $5x^2-(2x^2+9)=0$
$\Rightarrow 3x^2=9$
$\Rightarrow x=-\sqrt{3},\sqrt{3}$


Required area
$=2\underset{0}{\overset{\sqrt{3}}{\int }}((2x^2+9)-5x^2)dx$
$=2\underset{0}{\overset{\sqrt{3}}{\int }}(9-3x^2)dx$
$=2{[9x-x^3]}_0^{\sqrt{3}}$
$=2[9\sqrt{3}-3\sqrt{3}]$
$=12\sqrt{3}$ sq. units