Let $$P_1 : y=4x^2 \text{ and } P_2 : y=x^2 + 27$$ 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 $$y = c\alpha x, \alpha > 0 \text{ and } P_1,$$ then $$\alpha$$ is equal to:

Area between parabolas ($$A_1$$).

Find intersection: $$4x^2 = x^2 + 27 \implies 3x^2 = 27 \implies x = \pm 3$$.

$$A_1 = \int_{-3}^{3} (x^2 + 27 - 4x^2) dx = \int_{-3}^{3} (27 - 3x^2) dx = [27x - x^3]_{-3}^{3} = 54 - (-54) = 108$$

Area between line and parabola ($$A_2$$).

Intersection: $$4x^2 = \alpha x \implies x = 0, x = \alpha/4$$.

$$A_2 = \int_{0}^{\alpha/4} (\alpha x - 4x^2) dx = \left[ \frac{\alpha x^2}{2} - \frac{4x^3}{3} \right]_{0}^{\alpha/4} = \frac{\alpha^3}{32} - \frac{\alpha^3}{48} = \frac{\alpha^3}{96}$$

Set up the ratio.

$$A_1 = 6 \cdot A_2 \implies 108 = 6 \left( \frac{\alpha^3}{96} \right) \implies 108 = \frac{\alpha^3}{16}$$

$$\alpha^3 = 108 \times 16 = 1728 \implies \alpha = \sqrt[3]{1728} = 12$$.