Question

The area bounded by the parabola $y^2=4ax,$ latus rectum and x-axis is

• A. 0
• B. $\frac{4}{3}{a}^2$
• C. $\frac{2}{3}{a}^2$
• D. $\frac{a^2}{3}$

Answer 1

The correct option is A 0

The equation of the parabola is $y^2=4ax$ ...$\left(1\right)$

Let $O$ be the vertex,$S$ be the focus and ${LL}^{\prime }$ be the latus rectum of araboal.

The equation of latus rectum is $x=a$

Also , we know that parabola is symmetric about $x-$axis

$\therefore$ Required Area$=2\left(areaOSL\right)$

$=2{\int }_0^{a}y.dx$

$=2{\int }_0^{a}2\sqrt{a}\sqrt{x}.dx$

$=4\sqrt{a}{\int }_0^a\sqrt{x}.dx$

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

$=\frac{8\sqrt{a}}{3}{a}^{\frac{3}{2}}=\frac{8a^2}{3}$sq.units.