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
Show Answer
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 $=2 \int_0^{2}(x^{2}+6-\frac{5 x^{2}}{2}) d x=\int_0^{\frac{2 \alpha}{5}}(\alpha x-\frac{5 x^{2}}{2}) d x$
$\Rightarrow \int_0^{\frac{2 \alpha}{5}}(\alpha x-\frac{5 x^{2}}{2}) d x=16$
$\Rightarrow \alpha^{3}=600$
