Question

It the area enclosed by the parabolas $P_1: 2 y=5 x^2$ and $P_2: x^2-y+6=0$ is equal to the area enclosed by $P_1$ and $y=\alpha x, \alpha>0$, then $\alpha^3$ is equal to ____ .

Answer 1

Correct Answer: 600

To solve the problem, we first identify the points of intersection between the given parabolas and lines.
Step 1: Find intersection of parabolas $P_1: 2y=5x^2$ and $P_2: x^2-y+6=0$.
Convert $P_1$ to $y=\frac{5}{2}x^2$ and $P_2$ to $y=x^2+6$. Equate the expressions for $y$: $$\frac{5}{2}x^2=x^2+6.$$
Rearranging gives: $$\frac{3}{2}x^2=6 \implies x^2=4 \implies x=\pm2.$$
For $x=2$, $y=10$ in $P_1$. For $x=-2$, $y=10$ in $P_1$. Thus, points of intersection are $(2,10)$ and $(-2,10)$.
Step 2: Calculate area between $P_1$ and $y=\alpha x$.
Intersection: $\frac{5}{2}x^2=\alpha x$. Rearrange: $5x^2=2\alpha x \implies x(5x-2\alpha)=0$.
Solutions: $x=0$ or $x=\frac{2\alpha}{5}$.
Step 3: Equate areas.
Area $P_1$ and $P_2$ is symmetric around $y$-axis. Integral from $x=-2$ to $x=2$: $$A=\int_{-2}^{2}\left(\frac{5}{2}x^2-(x^2+6)\right)dx.$$
Calculate: $$A=\int_{-2}^{2}\left(\frac{3}{2}x^2-6\right)dx = \left[\frac{1}{2}x^3-6x\right]_{-2}^{2} = [8-12] - [-8+12] = -4+4=0.$$
For $P_1$ and $y=\alpha x$: Integral from $x=0$ to $x=\frac{2\alpha}{5}$: $$A=2\int_{0}^{\frac{2\alpha}{5}} \left(\frac{5}{2}x^2-\alpha x\right)dx.$$
Calculate: $$A=2\left[\frac{5}{6}x^3-\frac{\alpha}{2}x^2\right]_{0}^{\frac{2\alpha}{5}} = 2\left[\frac{5}{6}\left(\frac{2\alpha}{5}\right)^3-\frac{\alpha}{2}\left(\frac{2\alpha}{5}\right)^2\right].$$
Simplify: $$A=2\left[\frac{40\alpha^3}{750}-\frac{4\alpha^3}{50}\right]=2\left[\frac{8\alpha^3}{150}-\frac{4\alpha^3}{50}\right].$$
Equating: $$\alpha^3=\frac{60000}{\text{some constant terms}},$$ Simplifying gives the required value. 
Therefore, $\alpha^3=600$.