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Q.
Area lying between the parabola $y^{2} = 4ax$ and its latus rectum is
Application of Integrals
Solution:
We have, $y^{2} = 4ax$, parabola with vertex $(0, 0)$ and focus $(a, 0)$ and latus rectum $4a$.
Required area = area of shaded region
$A=2\int\limits_{0}^{a}2 \sqrt{a}\sqrt{x} dx$
$2\left[2\sqrt{a}x^{3/ 2}\times\frac{2}{3}\right]_{0}^{a}$
$=\frac{8}{3}a^{2}$ sq. units
