Question:

Step 1: Identify the equations of the given parabolas and line. The first parabola is $P_1: 2y = 5x^2$, which can be rewritten as $y = \frac{5}{2}x^2$. The second parabola is $P_2: x^2 - y + 6 = 0$, which can be rewritten as $y = x^2 + 6$. The line is $y = \alpha x$, with $\alpha > 0$. Step 2: Calculate the area enclosed by parabolas $P_1$ and $P_2$. First, find the points of intersection of $P_1$ and $P_2$ by setting their $y$-values equal: $$ \frac{5}{2}x^2 = x^2 + 6 $$ $$ \frac{5}{2}x^2 - x^2 = 6 $$ $$ \frac{3}{2}x^2 = 6 $$ $$ x^2 = 4 $$ $$ x = \pm 2 $$ The intersection points are at $x=-2$ and $x=2$. To determine which curve is above the other in the interval $[-2, 2]$, we can test a point, e.g., $x=0$: For $P_1$, $y = \frac{5}{2}(0)^2 = 0$. For $P_2$, $y = (0)^2 + 6 = 6$. Since $6 > 0$, $P_2$ is the upper curve ($y_{upper} = x^2+6$) and $P_1$ is the lower curve ($y_{lower} = \frac{5}{2}x^2$). The area $A_1$ is given by the integral of the difference between the upper and lower curves from $x=-2$ to $x=2$: $$ A_1 = \int_{-2}^{2} \left( (x^2+6) - \frac{5}{2}x^2 \right) dx $$ $$ A_1 = \int_{-2}^{2} \left( 6 - \frac{3}{2}x^2 \right) dx $$ Since the integrand is an even function and the limits are symmetric, we can write: $$ A_1 = 2 \int_{0}^{2} \left( 6 - \frac{3}{2}x^2 \right) dx $$ Now, perform the integration: $$ A_1 = 2 \left[ 6x - \frac{3}{2} \cdot \frac{x^3}{3} \right]_{0}^{2} $$ $$ A_1 = 2 \left[ 6x - \frac{1}{2}x^3 \right]_{0}^{2} $$ Evaluate the definite integral: $$ A_1 = 2 \left( \left( 6(2) - \frac{1}{2}(2)^3 \right) - \left( 6(0) - \frac{1}{2}(0)^3 \right) \right) $$ $$ A_1 = 2 \left( (12 - \frac{1}{2}(8)) - 0 \right) $$ $$ A_1 = 2 (12 - 4) $$ $$ A_1 = 2 (8) $$ $$ A_1 = 16 $$ Step 3: Calculate the area enclosed by parabola $P_1$ and the line $y=\alpha x$. First, find the points of intersection of $P_1$ and $y=\alpha x$ by setting their $y$-values equal: $$ \frac{5}{2}x^2 = \alpha x $$ $$ \frac{5}{2}x^2 - \alpha x = 0 $$ $$ x \left( \frac{5}{2}x - \alpha \right) = 0 $$ This gives two solutions for $x$: $x=0$ or $\frac{5}{2}x = \alpha$, which means $x = \frac{2\alpha}{5}$. Since $\alpha > 0$, the interval of integration is $\left[0, \frac{2\alpha}{5}\right]$. To determine which curve is above the other in this interval, we can test a point, e.g., $x = \frac{\alpha}{5}$: For the line, $y_{line} = \alpha \left(\frac{\alpha}{5}\right) = \frac{\alpha^2}{5}$. For $P_1$, $y_{P_1} = \frac{5}{2} \left(\frac{\alpha}{5}\right)^2 = \frac{5}{2} \frac{\alpha^2}{25} = \frac{\alpha^2}{10}$. Since $\alpha > 0$, $\frac{\alpha^2}{5} > \frac{\alpha^2}{10}$, so the line $y=\alpha x$ is the upper curve ($y_{upper} = \alpha x$) and $P_1$ is the lower curve ($y_{lower} = \frac{5}{2}x^2$). The area $A_2$ is given by the integral of the difference between the upper and lower curves from $x=0$ to $x=\frac{2\alpha}{5}$: $$ A_2 = \int_{0}^{\frac{2\alpha}{5}} \left( \alpha x - \frac{5}{2}x^2 \right) dx $$ Now, perform the integration: $$ A_2 = \left[ \frac{\alpha x^2}{2} - \frac{5}{2} \cdot \frac{x^3}{3} \right]_{0}^{\frac{2\alpha}{5}} $$ $$ A_2 = \left[ \frac{\alpha x^2}{2} - \frac{5}{6}x^3 \right]_{0}^{\frac{2\alpha}{5}} $$ Evaluate the definite integral: $$ A_2 = \left( \frac{\alpha}{2} \left(\frac{2\alpha}{5}\right)^2 - \frac{5}{6} \left(\frac{2\alpha}{5}\right)^3 \right) - (0) $$ $$ A_2 = \frac{\alpha}{2} \left(\frac{4\alpha^2}{25}\right) - \frac{5}{6} \left(\frac{8\alpha^3}{125}\right) $$ $$ A_2 = \frac{4\alpha^3}{50} - \frac{40\alpha^3}{750} $$ Simplify the fractions: $$ A_2 = \frac{2\alpha^3}{25} - \frac{4\alpha^3}{75} $$ Combine the terms by finding a common denominator (75): $$ A_2 = \frac{3 \cdot 2\alpha^3}{75} - \frac{4\alpha^3}{75} $$ $$ A_2 = \frac{6\alpha^3 - 4\alpha^3}{75} $$ $$ A_2 = \frac{2\alpha^3}{75} $$ Step 4: Equate the two areas and solve for $\alpha^3$. According to the problem statement, the area $A_1$ is equal to the area $A_2$: $$ A_1 = A_2 $$ $$ 16 = \frac{2\alpha^3}{75} $$ Now, solve for $\alpha^3$: $$ 2\alpha^3 = 16 \times 75 $$ $$ \alpha^3 = \frac{16 \times 75}{2} $$ $$ \alpha^3 = 8 \times 75 $$ $$ \alpha^3 = 600 $$ The final answer is $\boxed{600}$.