Let P_1 : y = 4x^2 and P_2 : y = x^2 + 27 be two parabolas.

Question

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 $P_1$,...

Accepted Answer

Calculate the Area ${A}_{1}$ between ${P}_{1}$ and ${P}_{2}$

The two parabolas are given by:

${P}_{1}:y=4{x}^{2}$

${P}_{2}:y={x}^{2}+27$

Find the intersection points: Setting the equations equal to each other:

$4{x}^{2}={x}^{2}+27$

$\Rightarrow$$3{x}^{2}=27$

$\Rightarrow$ ${x}^{2}=9\Rightarrow x=\pm 3$

$\begin{gathered} \ 4{x}^{2}={x}^{2}+27 \ 3{x}^{2}=27\, ,\, {x}^{2}=9\Rightarrow x=\pm 3 $

Calculate the area: On the interval $[-3,3]$, the parabola ${P}_{2}$ lies above ${P}_{1}$. The area ${A}_{1}$ is:

$\begin{gathered} {A}_{1}={\int }_{3}^{-3}({x}^{2}+27-4{x}^{2})dx={\int }_{3}^{-3}(27-3{x}^{2})dx \ {A}_{1}={[27x-{x}^{3}]}_{3}^{-3} \ {A}_{1}=(27(3)-{3}^{3})-(27(-3)-(-3{)}^{3}) \ {A}_{1}=(81-27)-(-81+27)=54-(-54)=108 \end{gathered}$

Calculate the Area ${A}_{2}$ between ${P}_{1}$ and the line $y=\alpha x$

Find the intersection points:

$4{x}^{2}=\alpha x\implies x(4x-\alpha )=0$

The points are $x=0$ and $x=\frac{\alpha }{4}$.

Calculate the area: On the interval $[0,\frac{\alpha }{4}]$, the line $y=\alpha x$ lies above $y=4{x}^{2}$. The area ${A}_{2}$ is:

$ {A}_{2} ={\int }_{\alpha /4}^{0}(\alpha x-4{x}^{2})dx \ {A}_{2} ={[\frac{\alpha {x}^{2}}{2}-\frac{4{x}^{3}}{3}]}_{\alpha /4}^{0} \ {A}_{2} =\frac{\alpha }{2}{(\frac{\alpha }{4})}^{2}-\frac{4}{3}{(\frac{\alpha }{4})}^{3} \ {A}_{2} =\frac{{\alpha }^{3}}{32}-\frac{4{\alpha }^{3}}{192}=\frac{{\alpha }^{3}}{32}-\frac{{\alpha }^{3}}{48}=\frac{3{\alpha }^{3}-2{\alpha }^{3}}{96}=\frac{{\alpha }^{3}}{96} \end{gathered}$

The problem states that ${A}_{1}$ is six times ${A}_{2}$ :

$\begin{gathered} {A}_{1}=6	imes {A}_{2} \ 108=6	imes (\frac{{\alpha }^{3}}{96}) \ 108=\frac{{\alpha }^{3}}{16} \ {\alpha }^{3}=108	imes 16=1728 \ \alpha =\sqrt31728 \end{gathered}$

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