JEE MAIN - Mathematics (2023 - 25th January Morning Shift - No. 22)
If the area enclosed by the parabolas $$\mathrm{P_1:2y=5x^2}$$ and $$\mathrm{P_2:x^2-y+6=0}$$ is equal to the area enclosed by $$\mathrm{P_1}$$ and $$\mathrm{y=\alpha x,\alpha > 0}$$, then $$\alpha^3$$ is equal to ____________.
Answer
600
Explanation
$x^{2}+6=\frac{5}{2} x^{2} \Rightarrow x=\pm 2$
$$ \begin{aligned} & \text { Area between } P_{1} \text { and } P_{2} \quad \text { [Say } \left.A_{1}\right] \\\\ & =\int\limits_{-2}^{2}\left(x^{2}+6\right)-\frac{5}{2} x^{2} d x \\\\ & =2 \int\limits_{0}^{2}\left(6-\frac{3}{2} x^{2}\right) d x=2\left[6 x-\frac{x^{3}}{2}\right]_{0}^{2}=16 \end{aligned} $$
$$ \begin{aligned} & a x=\frac{5}{2} x^2 \Rightarrow x=0, \frac{2 a}{5} \\\\ & \text { Area between } P_1 \text { and } y=a x \quad \text { [Say } A_2 \text { ] } \\\\ & =\int\limits_0^{\frac{2 \alpha}{5}} a x-\frac{5}{2} x^2 d x \\\\ & \left.=\frac{a x^2}{2}-\frac{5}{6} x^3\right]_0^{\frac{2 a}{5}}: \frac{2 a^3}{75} \\\\ & A_1=A_2 \Rightarrow \frac{2 a^3}{75}=16 \\\\ & a^3=600 \end{aligned} $$
$$ \begin{aligned} & \text { Area between } P_{1} \text { and } P_{2} \quad \text { [Say } \left.A_{1}\right] \\\\ & =\int\limits_{-2}^{2}\left(x^{2}+6\right)-\frac{5}{2} x^{2} d x \\\\ & =2 \int\limits_{0}^{2}\left(6-\frac{3}{2} x^{2}\right) d x=2\left[6 x-\frac{x^{3}}{2}\right]_{0}^{2}=16 \end{aligned} $$
$$ \begin{aligned} & a x=\frac{5}{2} x^2 \Rightarrow x=0, \frac{2 a}{5} \\\\ & \text { Area between } P_1 \text { and } y=a x \quad \text { [Say } A_2 \text { ] } \\\\ & =\int\limits_0^{\frac{2 \alpha}{5}} a x-\frac{5}{2} x^2 d x \\\\ & \left.=\frac{a x^2}{2}-\frac{5}{6} x^3\right]_0^{\frac{2 a}{5}}: \frac{2 a^3}{75} \\\\ & A_1=A_2 \Rightarrow \frac{2 a^3}{75}=16 \\\\ & a^3=600 \end{aligned} $$
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