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

12 \sqrt{3}

sq. units

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6 \sqrt{3}

sq. units

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8 \sqrt{3}

sq. units

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4 \sqrt{3}

sq. units

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Solution

The correct option is A $12 \sqrt{3}$ sq. units
Given $5 x^{2} - y = 0$
$2 x^{2} - y + 9 = 0$
Eliminating $y$ , we get $5 x^{2} - (2 x^{2} + 9) = 0$
$\Rightarrow 3 x^{2} = 9$
$\Rightarrow x = - \sqrt{3}, \sqrt{3}$

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

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