Question

Draw a rough sketch of the curve and find the area of the region bounded by curve $y^2=8x$ and the line $x=2$.

Answer 1

Here we need to find the area of the region bounded by the curve $y^2=8x$ and the line $x=2$

$y^2=8x$ is a parabola that opens right and its focus is $(4a=8\Rightarrow a=2)\Rightarrow (2,0)$

Hence $x=2$ is the latus rectum of the parabola.

The required area is the shaded portion shown in the figure.

The required area is the region bounded between the latus rectum.Since it is symmetrical about $x-axis$ which includes $IandIV$ quadrant.

Hence $A=2\times {\int }_0^2\sqrt{8x}dx$

$=2\sqrt{8}{\int }_0^2\sqrt{x}dx$

$=2\sqrt{8}{\left[\frac{x^{3/2}}{3/2}\right]}_0^2$

$=\frac{4\sqrt{8}}{3}{\left[x^{3/2}\right]}_0^2$

$=\frac{8\sqrt{2}}{3}(2\sqrt{2}-0)$

$A=\frac{32}{3}sq.units$