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Q.
The area of the region bounded by the parabola $y^2 = 4ax$ and its latus rectum is:
Application of Integrals
Solution:
Given parabola is $y^2 = 4ax$
$\Rightarrow \, y = 2 \sqrt{a} \sqrt{x}$
Vertex of this parabola is (0, 0) and focus is (a, 0)
Axis of this parabola is x-axis and is symmetric about x-axis.
Latus-rectum is LL' which is perpendicular to x-axis and passing through focus.
$\therefore $ Required area is area of the region OLL' = 2 $\times$ area of region OSL
$2 \int^{a}_{0} ydx = 2 \int^{a}_{0} 2\sqrt{a} \sqrt{x} dx $
$= 4 \sqrt{a } \left(\frac{2}{3} x^{\frac{3}{2}}\right)^{a}_{0} = \frac{8\sqrt{a}}{3} \left[a^{\frac{3}{2}}\right] = \frac{8}{3} a^{2} $