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Q.
If the area bounded by $y^{2}=4ax$ and $x^{2}=4ay$ is $\frac{64}{3}$ square units, then the positive value of $a$ is
NTA AbhyasNTA Abhyas 2020Application of Integrals
Solution:
Point of intersection of the two given curves are $\left(0,0\right)$ and $\left(4 a , 4 a\right)$
The area bounded $=2\displaystyle \int _{0}^{4 a} \left(x - \frac{x^{2}}{4 a}\right) d x$
$=2\left[\frac{x^{2}}{2} - \frac{x^{3}}{12 a}\right]_{0}^{4 a}$
$=2\left[8 a^{2} - \frac{16 a^{2}}{3}\right]=\frac{16 a^{2}}{3}$ sq. units
Given, $\frac{16 a^{2}}{3}=\frac{64}{3}$
Hence, the positive value of $a=2$
