Let $P_{1}:y = 4x^{2}$ and $P_{2}:y = x^{2} + 27$ be two parabolas. If the area of the bounded regio

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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:

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Area bounded between $P_{1}\& P_{2}$ is$$\int_{- 3}^{3}\mspace{2mu}\left( \left( x^{2} + 27 ight) - \left( 4x^{2} ight) ight)dx $$(P.O.I. of $P_{1}\& P_{2}$ is $x = \pm 3$ )$${= 2\int_{0}^{3}\mspace{2mu}\left( 27 - 3x^{2} ight)dx = 2\left\lbrack 27x - x^{3} ightbrack_{0}^{3} }{= 2\lbrack 81 - 27brack = 108 }$$∴ Area bounded between $P_{1}\&	ext{ }L$ is 18 sq. units(Area between $x^{2} = 4$ ay & line $x = my$ ) is $\frac{8a^{2}}{3{	ext{ }m}^{3}}$∴ Area between $x^{2} = \frac{y}{4}\& x = \frac{y}{\alpha}$ is$${\frac{8 \cdot \left( \frac{1}{16} ight)^{2}}{3 \cdot \left( \frac{1}{\alpha} ight)^{3}} = 18 }{\Rightarrow \frac{\frac{8}{16.16}}{\frac{3}{\alpha^{3}}} = 18 \Rightarrow \alpha^{3} = 2^{6}{.3}^{3} }{\Rightarrow \alpha = 12}$$

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