Question

It the area enclosed by the parabolas $P_1: 2 y=5 x^2$ and $P_2: x^2-y+6=0$ is equal to the area enclosed by $P_1$ and $y=\alpha x, \alpha>0$, then $\alpha^3$ is equal to_____

Answer 1

$ \text { Area }=2 \int\limits^2\left(x^2+6-\frac{5 x^2}{2}\right) d x=\int\limits_0^{\frac{2 \alpha}{5}}\left(\alpha x-\frac{5 x^2}{2}\right) d x$
$ \Rightarrow \int\limits_0^{\frac{2 \alpha}{5}}\left(\alpha x-\frac{5 x^2}{2}\right) d x=16 $
$ \Rightarrow \alpha^3=600$