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

4 \sqrt{3}

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6 \sqrt{3}

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12 \sqrt{3}

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18 \sqrt{3}

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Solution

The correct option is C $12 \sqrt{3}$

By solving $5 x^{2} - y = 0, 2 x^{2} - y + 9 = 0 w e g e t, x = \pm \sqrt{3}$
Area = $2 \int_{0}^{\sqrt{3}} (9 - 3 x^{2}) d x = 2 [9 x - x^{3}]_{0}^{\sqrt{3}} = 12 \sqrt{3}$ sq. unit

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Similar questions

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

5 x^{2} - y = 0

2 x^{2} - y + 9 = 0

5 x^{2} - y = 0

2 x^{2} - y + 9 = 0

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

5 x^{2} - y = 0

2 x^{2} - y + 9 = 0

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