If the area enclosed by the parabolas P_1:2y=5x^2 and P_2:x^2-y+6=0 is equal to the area e…

Question

If the area enclosed by the parabolas $P_1:2y=5x^2$ 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: 600

Accepted Answer

Intersection of $P_1,P_2$ ($y=x^2+6$): $2(x^2+6)=5x^2\Rightarrow x^2=4\Rightarrow x=\pm2$.

$$\text{Area}(P_1,P_2)=\int_{-2}^{2}\left[6-\frac{3x^2}{2}\right]dx=2\left[6x-\frac{x^3}{2}\right]_0^2=2(8)=16.$$

Intersection of $P_1$ and $y=\alpha x$: $x=0$ or $x=\dfrac{2\alpha}{5}$.

$$\text{Area}=\int_0^{2\alpha/5}\left[\alpha x-\frac{5x^2}{2}\right]dx = \frac{2\alpha^3}{75}=16 \;\Rightarrow\; \alpha^3=600.$$

Solutions · 1