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

Correct Answer: 600

The correct answer is 600.

$\text{Area}=2{\int }_0^2\left(x^2+6-\frac{5x^2}{2}\right)dx=\underset{0}{\overset{\frac{2\alpha }{5}}{\int }}\left(\alpha x-\frac{5x^2}{2}\right)dx$
$\Rightarrow \underset{0}{\overset{\frac{2\alpha }{5}}{\int }}\left(\alpha x-\frac{5x^2}{2}\right)dx=16$
$\Rightarrow {\alpha }^3=600$