Q. 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 :

Correct Answer: 12

Explanation

First find the points of intersection of the parabolas $P_1$ and $P_2$.

$4x^2 = x^2 + 27 \Rightarrow 3x^2 = 27 \Rightarrow x = \pm 3$

Area enclosed between $P_2$ and $P_1$ is

$\displaystyle \int_{-3}^{3} \left[(x^2+27) - 4x^2\right] dx = \int_{-3}^{3} (27 - 3x^2)\,dx$

$= 2\int_{0}^{3} (27 - 3x^2)\,dx = 2\left[27x - x^3\right]_0^3 = 2(81 - 27)=108$

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Now consider the area enclosed between the line $y=\alpha x$ and the parabola $y=4x^2$.

$4x^2 = \alpha x \Rightarrow x(4x-\alpha)=0$

So the points of intersection are $x=0$ and $x=\alpha/4$.

Area enclosed is

$\displaystyle \int_{0}^{\alpha/4} (\alpha x - 4x^2)\,dx = \left[\frac{\alpha x^2}{2} - \frac{4x^3}{3}\right]_0^{\alpha/4}$

$= \frac{\alpha^3}{32} - \frac{\alpha^3}{48} = \frac{\alpha^3}{96}$

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Given that the first area is six times the second:

$108 = 6 \times \frac{\alpha^3}{96}$

$\alpha^3 = 1728 \Rightarrow \alpha = 12$

Hence, the value of $\alpha$ is

12

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