Area Under Curves Question 3

Question 3 - 25 January - Shift 1

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

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Answer: 600

Solution:

Formula: Area between two curves - Area enclosed between two curves intersecting at two different points

Graph between $P_1$ and $P_2$

Abscissa of point of intersection of $2 y=5 x^{2}$ and $y=x^{2}+6$ is $\pm 2$

Graph between $P_1$ and $y=\alpha x, \alpha>0$

Area $=2 \int_0^{2}(x^{2}+6-\frac{5 x^{2}}{2}) d x=\int_0^{\frac{2 \alpha}{5}}(\alpha x-\frac{5 x^{2}}{2}) d x$

$\Rightarrow \int_0^{\frac{2 \alpha}{5}}(\alpha x-\frac{5 x^{2}}{2}) d x=16$

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