Question

The area 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
$5x^2-(2x^2+9)=0\Rightarrow x=-\sqrt{3},\sqrt{3}$

$\therefore$ required area
$=2{\int }_0^{\sqrt{3}}\left(\right(2x^2+9)-5x^2)dx=2{\int }_0^{\sqrt{3}}(9-3x^2)dx$
$=12\sqrt{3}$ sq. units