Area bounded between $P_1$ \& $P_2$ is $$ \begin{aligned} & \int_{-3}^3\left(\left(x^2+27\right)-\left(4 x^2\right)\right) d x \\ & \left.\quad \text { (P.O.I. of } P_1 \text { \& } P_2 \text { is } x= \pm 3\right) \\ & =2 \int_0^3\left(27-3 x^2\right) d x=2\left[27 x-x^3\right]_0^3 \\ & =2[81-27]=108 \end{aligned} $$ ∴ Area bounded between $\mathrm{P}_1$ \& L is 18 sq. units (Area between $\mathrm{x}^2=4 \mathrm{ay}$ \& line $\mathrm{x}=\mathrm{my}$ ) is $\frac{8 \mathrm{a}^2}{3 \mathrm{~m}^3}$ ∴ Area between $\mathrm{x}^2=\frac{\mathrm{y}}{4} \& \mathrm{x}=\frac{\mathrm{y}}{\alpha}$ is $$ \begin{aligned} & \frac{8 \cdot\left(\frac{1}{16}\right)^2}{3 \cdot\left(\frac{1}{\alpha}\right)^3}=18 \\ & \Rightarrow \frac{\frac{8}{16 \cdot 16}}{\frac{3}{\alpha^3}}=18 \Rightarrow \alpha^3=2^6 \cdot 3^3 \\ & \Rightarrow \alpha=12 \end{aligned} $$