Question

Find the area bounded by the curves $y^2=4a^2(x-1)$ and the line $x=1,y=4a$.

Answer 1

The equation of parabola is $y^2=4a(x-1)$ ......$\left(1\right)$

When $y=4a,$ from $\left(1\right)16a^2=4a(x-1)$

$\Rightarrow x-1=4a\Rightarrow x=4a+1$

Required area$={\int }_{x=1}^{x=4a+1}ydx$

$={\int }_{x=1}^{x=4a+1}2\sqrt{a}\sqrt{x-1}dx\because of\left(1\right)$

$=2\sqrt{a}{\int }_1^{4a+1}\sqrt{x-1}dx$

$=2\sqrt{a}{\left[\frac{{(x-1)}^{\frac{1}{2}+1}}{\frac{1}{2}+1}\right]}_1^{4a+1}$

$=2\sqrt{a}{\left[\frac{2{(x-1)}^{\frac{3}{2}}}{3}\right]}_1^{4a+1}$

$=2\sqrt{a}[\frac{2{(4a+1-1)}^{\frac{3}{2}}}{3}-\frac{2{(1-1)}^{\frac{3}{2}}}{3}]$

$=2\sqrt{a}[\frac{2{\left(4a\right)}^{\frac{3}{2}}}{3}-0]$

$=\frac{4}{3}\times 8\times a^{\frac{1}{2}+\frac{3}{2}}$

$=\frac{32}{3}{a}^2$sq.units.