Question

​The area of the bounded by the parabola y² = 4ax and its latusrectum is ____________.

Answer 1

To find: area of the region bounded by the parabola y² = 4ax and its latusrectum

Latusrectum of a parabola is x = 4a



The required area of the region OCBO = 2(Area of the region OABO)
Thus,
$$\begin{gathered} \mathrm{Required}\mathrm{area}=2{\int }_0^{4a}ydx \\ =2{\int }_0^{4a}\sqrt{4ax}dx \\ =2{\int }_0^{4a}2\sqrt{a}\sqrt{x}dx \\ =4\sqrt{a}{\left(\frac{x^{{\displaystyle \frac{3}{2}}}}{{\displaystyle \frac{3}{2}}}\right)}_0^{4a} \\ =4\sqrt{a}\times \frac{2}{3}\left[{\left(4a\right)}^{\frac{3}{2}}-0\right] \\ =\frac{8}{3}\sqrt{a}\left[8{\left(a\right)}^{\frac{3}{2}}\right] \\ =\frac{64}{3}{a}^{\frac{3}{2}+\frac{1}{2}} \\ =\frac{64}{3}{a}^{\frac{4}{2}} \\ =\frac{64}{3}{a}^2 \\ \mathrm{Thus},\mathrm{Area}=\frac{64}{3}{a}^2\mathrm{sq}.\mathrm{units} \end{gathered}$$


Hence, the area of the region bounded by the parabola y² = 4ax and its latusrectum is $\underline{\frac{64}{3}{a}^2\mathrm{sq}.\mathrm{units}}.$