If 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…

Given:

$P_1:2y=5x^2$

$P_2:x^2-y+6=0$

On solving both equations, we have

$2y=5\left(y-6\right)$

$\Rightarrow 3y=30$

$\Rightarrow y=10$

$\Rightarrow x=\pm 2$

So, points of intersection are $\left(2,10\right) \& \left(-2,10\right)$.

Also, solving $P_1$ and $y=\alpha x$, we have

$x=0, \frac{2\alpha }{5}$

According to question $2{\int }_0^2\left(x^2+6-\frac{5x^2}{2}\right)dx={\int }_0^{\frac{2\alpha }{5}}\left(\alpha x-\frac{5x^2}{2}\right)dx$

$\Rightarrow 2{\left[\frac{x^3}{3}+6x-\frac{5x^3}{6}\right]}_0^2={\left[\frac{\alpha x^2}{2}-\frac{5x^3}{6}\right]}_0^{\frac{2\alpha }{5}}$

$\Rightarrow 2\left(12+\frac{8}{3}-\frac{20}{3}\right)=\left[\frac{2{\alpha }^3}{25}-\frac{8{\alpha }^3}{6\times 25}\right]$

$\Rightarrow \frac{2{\alpha }^3}{75}=16$

$\Rightarrow {\alpha }^3=600$