Question

Find the area of the parabola $y^2=4ax$ bounded by its latus rectum.

Answer 1

Given: $y^2=4ax$

Latus rectum of this parabola is line
$x=a$

Draw figure

$y^2=4ax$

$\to y=\pm \sqrt{4ax}$

$\to y=\pm 2\sqrt{ax}$

Area required

$=2\times \text{Area OSLO}$

$-2\times {\int }_0^{a}ydx$

$-2\times {\int }_0^{a}2\sqrt{ax}dx$

$=4\sqrt{a}{\int }_0^a(x{)}^{\frac{1}{2}}dx$

$=4\sqrt{a}{\left[\frac{x^{\frac{3}{2}}}{\frac{3}{2}}\right]}_0^a$

$=4\sqrt{a}\times \frac{2}{3}{\left[x^{\frac{3}{2}}\right]}_0^a$

$=\frac{8}{3}\sqrt{a}[a^{\frac{3}{2}}-0]$

$=\frac{8}{3}{a}^{\frac{1}{2}}\left(a^{\frac{3}{2}}\right)$

$=\frac{8a^2}{3}$

Final answer:
Therefore, required Area =$\frac{8a^2}{3}$ square units.