Question

The area bounded by the parabola y² = 4ax, latusrectum 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

$\left(\mathrm{b}\right)\frac{4}{3}{a}^2$


Clearly, the latusrectum passes x-axis through the point D(a, 0).
Therefore, the required area ABCD,
$$\begin{gathered} A={\int }_0^{a}y\mathit{d}x\mathit{}\mathit{}\left(\mathrm{Where},y=2\sqrt{ax}\right) \\ ={\int }_0^{1}2\sqrt{ax}\mathit{d}x \\ ={\left[\frac{4\sqrt{a}}{3}{\left(x\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right]}_0^a \\ =\left[\frac{4\sqrt{a}}{3}{\left(a\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right]-\left[\frac{4\sqrt{a}}{3}{\left(0\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right] \\ =\frac{4}{3}{a}^2\mathrm{square}\mathrm{units} \end{gathered}$$