Let $P_{1}: y=4 x^{2}$ 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
  A

  12
• B

  6
• C

  15
• D

  8

Solution

Area bounded between $P_1$ & $P_2$ is
$\int_{-3}^{3} ((x^2 + 27) - (4x^2)) \, dx$
(P.O.I. of $P_1$ & $P_2$ is $x = \pm 3$)
$= 2\int_{0}^{3} (27 - 3x^2) \, dx = 2[27x - x^3]_{0}^{3}$
$= 2[81 - 27] = 108$
$\therefore$ Area bounded between $P_1$ & L is 18 sq. units
(Area between $x^2 = 4ay$ & line $x = my$) is $\dfrac{8a^2}{3m^3}$
$\therefore$ Area between $x^2 = \dfrac{y}{4}$ & $x = \dfrac{y}{\alpha}$ is
$\dfrac{8 \cdot \left(\dfrac{1}{16}\right)^2}{3 \cdot \left(\dfrac{1}{\alpha}\right)^3} = 18$
$\Rightarrow \dfrac{\dfrac{8}{16 \cdot 16}}{\dfrac{3}{\alpha^3}} = 18 \Rightarrow \alpha^3 = 2^6 \cdot 3^3$
$\Rightarrow \alpha = 12$

Hence, the correct answer is option: 'A'.