Let P_{1}: y = 4x^{2} and P_{2}: y = x^{2} + | vedclass

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(C) The points of intersection of $P_{1}: y = 4x^{2}$ and $P_{2}: y = x^{2} + 27$ are found by setting $4x^{2} = x^{2} + 27$, which gives $3x^{2} = 27

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$12$

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(C) The points of intersection of $P_{1}: y = 4x^{2}$ and $P_{2}: y = x^{2} + 27$ are found by setting $4x^{2} = x^{2} + 27$, which gives $3x^{2} = 27$, so $x = \pm 3$.The area $A_{1}$ bounded between $P_{1}$ and $P_{2}$ is $\int_{-3}^{3} ((x^{2} + 27) - 4x^{2}) dx = \int_{-3}^{3} (27 - 3x^{2}) dx = 2 \int_{0}^{3} (27 - 3x^{2}) dx = 2 [27x - x^{3}]_{0}^{3} = 2(81 - 27) = 108$ sq. units.According to the problem, the area $A_{2}$ bounded between $P_{1}$ and the line $y = \alpha x$ is $A_{2} = \frac{A_{1}}{6} = \frac{108}{6} = 18$ sq. units.The line $y = \alpha x$ intersects $P_{1}: y = 4x^{2}$ at $4x^{2} = \alpha x$, so $x(4x - \alpha) = 0$, giving $x = 0$ and $x = \frac{\alpha}{4}$.The area $A_{2} = \int_{0}^{\alpha/4} (\alpha x - 4x^{2}) dx = [\frac{\alpha x^{2}}{2} - \frac{4x^{3}}{3}]_{0}^{\alpha/4} = \frac{\alpha}{2}(\frac{\alpha^{2}}{16}) - \frac{4}{3}(\frac{\alpha^{3}}{64}) = \frac{\alpha^{3}}{32} - \frac{\alpha^{3}}{48} = \frac{3\alpha^{3} - 2\alpha^{3}}{96} = \frac{\alpha^{3}}{96}$.Setting $A_{2} = 18$, we get $\frac{\alpha^{3}}{96} = 18$, so $\alpha^{3} = 18 	imes 96 = 1728$.Thus, $\alpha = \sqrt[3]{1728} = 12$.

Let $P_{1}: y = 4x^{2}$ and $P_{2}: y = x^{2} + 27$ be two parabolas. If the area of the bounded region enclosed between $P_{1}$ and $P_{2}$ is six times the area of the bounded region enclosed between the line $y = \alpha x, \alpha > 0$ and $P_{1}$, then $\alpha$ is equal to:

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