Question

The area bounded by the parabola y² = 8x, the x-axis and the latusrectum is
(a) $\frac{16}{3}$

(b) $\frac{23}{3}$

(c) $\frac{32}{3}$

(d) $\frac{16\sqrt{2}}{3}$

Answer 1

$\left(\mathrm{a}\right)\frac{16}{3}$




y² = 8x represents a parabola opening side ways , with vertex at O(0, 0) and Focus at B(2, 0)

Thus AA' represents the latus rectum of the parabola.

The points of intersection of the parabola and latus rectum are A(2, 4) and A'(2, −4)

Area bound by curve , x-axis and latus rectum is the area OABO,

$$\begin{gathered} \mathrm{The}\mathrm{approximating}\mathrm{rectangle}\mathrm{of}\mathrm{width}=dx\mathrm{and}length=\left|y\right|\mathrm{ha}sarea=\left|y\right|dx,\mathrm{and}\mathrm{moves}\mathrm{from}x=0\mathrm{to}x=2 \\ \mathrm{area}\left(OABO\right)={\int }_0^2\left|y\right|dx \\ ={\int }_0^{2}ydx\left\{y>0,\Rightarrow \left|y\right|=y\right\} \\ ={\int }_0^2\sqrt{8x}dx \\ =2\sqrt{2}{\int }_0^2\sqrt{x}dx \\ =2\sqrt{2}{\left[\frac{x^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}\right]}_0^2 \\ =2\sqrt{2}\times \frac{2}{3}\left(2^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-0\right) \\ =4\frac{\sqrt{2}}{3}\times 2\sqrt{2} \\ =\frac{16}{3}\mathrm{sq}\mathrm{units} \end{gathered}$$