Step 1: Calculate Area Between Parabolas $P_1$ and $P_2$
Find the intersection points of $P_1: y = 4x^2$ and $P_2: y = x^2 + 27$. $4x^2 = x^2 + 27$ $3x^2 = 27$ $x^2 = 9 \implies x = \pm 3$
The area, Area$_1$, is the integral of the difference between the upper curve ($P_2$) and the lower curve ($P_1$) from $x=-3$ to $x=3$. $Area$_1 = \int_{-3}^{3} ((x^2 + 27) - 4x^2) dx$ $Area$_1 = \int_{-3}^{3} (27 - 3x^2) dx$ Evaluating the integral: $Area$_1 = [27x - x^3]_{-3}^{3}$ $Area$_1 = (27(3) - 3^3) - (27(-3) - (-3)^3)$ $Area$_1 = (81 - 27) - (-81 + 27) = 54 - (-54) = 108$
Step 2: Calculate Area Between Line $y = \alpha x$ and Parabola $P_1$
Find the intersection points of the line $y = \alpha x$ ($\alpha > 0$) and parabola $P_1: y = 4x^2$. $\alpha x = 4x^2$ $x(4x - \alpha) = 0 \implies x = 0 \text{ or } x = \frac{\alpha}{4}$
The area, Area$_2$, is the integral from $x=0$ to $x=\frac{\alpha}{4}$. The line $y = \alpha x$ is the upper curve. $Area$_2 = \int_{0}^{\alpha/4} (\alpha x - 4x^2) dx$ Evaluating the integral: $Area$_2 = [\frac{\alpha x^2}{2} - \frac{4x^3}{3}]_{0}^{\alpha/4}$ $Area$_2 = \frac{\alpha (\alpha/4)^2}{2} - \frac{4(\alpha/4)^3}{3}$ $Area$_2 = \frac{\alpha^3}{32} - \frac{\alpha^3}{48}$ $Area$_2 = \frac{3\alpha^3 - 2\alpha^3}{96} = \frac{\alpha^3}{96}$
Step 3: Relate Areas and Solve for $\alpha$
Given the condition: Area$_1 = 6 \times Area$_2$. $108 = 6 \times \frac{\alpha^3}{96}$ Simplify the equation: $108 = \frac{\alpha^3}{16}$ Solve for $\alpha^3$: $\alpha^3 = 108 \times 16 = 1728$ Take the cube root: $\alpha = \sqrt[3]{1728} = 12$
Similar Questions
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Important Questions from Calculus
Let a line passing through the point $(4, 1, 0)$ intersect the line $L_1: \frac{x-1}{2} = \frac{y-2}{3} = \frac{z-3}{4}$ at the point $A(\alpha, \beta, \gamma)$ and the line $L_2: x-6=y=-z+4$ at the point $B(a, b, c)$. Then \(\begin{bmatrix} 1 & 0 & 1 \\ \alpha & \beta & \gamma \\ a & b & c \end{bmatrix}\) is equal to
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