If the area enclosed by the parabolas P _ 1 : 2y = 5 x ^ 2 and P _...

If the area enclosed by the parabolas

{P}_{1}: 2y = 5{x}^{2}

and

{P}_{2}: {x}^{2} - y + 6 = 0

is equal to the area enclosed by

{P}_{1}

and

y = \alphax, \alpha > 0

, then

{\alpha}^{3}

is equal to _____ . If the area enclosed by the parabolas P _ 1 : 2y = 5 x ^ 2 and P _ 2 : x ^ 2 - y + 6 = 0 is equal to the area enclosed

Correct answer

Step-by-step solution

Given:

${P}_{1}: 2y = 5{x}^{2}$

${P}_{2}: {x}^{2} - y + 6 = 0$

On solving both equations, we have

$2y = 5\left(y - 6\right)$ ⇒ 3y = 30 $$ ⇒ y = 10 $$ ⇒ x = \pm 2

So, points $of intersection are \left(2, 10\right) \text{ and } \left(- 2, 10\right).<img alt="" src="https://cdn-question-pool.Quantrex.app/pyq/jee_main/9203ecdb-dd8d-4202-8d3d-a8d66610a327-9894325_re.PNG" style="width: 303px; height: 202px;" />Also, solving {P}_{1} and$ y = \alphax $, we have$ x = 0, $\dfrac{2\alpha}{5}<img alt="" src="https://cdn-question-pool.Quantrex.app/pyq/jee_main/3b595a5e-541f-492b-8b4a-b8e550e7a052-9894325_1_re.PNG" style="width: 329px; height: 230px;" />According$ to question2{∫}_{0}^{2} $\left({x}^{2} + 6 - \dfrac{5{x}^{2}}{2}\right)$ dx = {∫}_{0}^ ${\dfrac{2\alpha}{5}}$ \left(\alphax - \dfrac{5{x}^{2}}{2}\right) $dx$

$⇒ 2{\left[\dfrac{{x}^{3}}{3} + 6x - \dfrac{5{x}^{3}}{6}\right]}_{0}^{2} = {\left[\dfrac{\alpha{x}^{2}}{2} - \dfrac{5{x}^{3}}{6}\right]$ }_{0}^ ${\dfrac{2\alpha}{5}}$ ⇒ 2 \left(12 + \dfrac{8}{3} - \dfrac{20}{3}\right) $= \left[\dfrac{2{\alpha}^{3}}{25} - \dfrac{8{\alpha}^{3}}{6 \times 25}\right]$

$⇒ \dfrac{2{\alpha}^{3}}{75} = 16$

$⇒ {\alpha}^{3} = 600$

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