Question

The area bounded by the curve $x=a{t}^2,y=2at$ and x-axis is $1\le t\le 3$ is

• A. $26a^2$
• B. $8a^2$
• C. $\frac{26a^2}{3}$
• D. $\frac{104a^2}{3}$

Answer 1

The correct option is D $\frac{104a^2}{3}$

Eliminating $t$, we ontain the equation of the curve as $y^2=4ax$

$t=1\Rightarrow x=a$ and $t=3\Rightarrow x=9a$

Hence required area

$={\int }_a^{9a}ydx={\int }_a^{9a}2\sqrt{a}\sqrt{x}dx$

$=2\sqrt{a}{\left[\frac{x^{3/2}}{3/2}\right]}_a^{9a}=\frac{4}{3}\sqrt{2}[{\left(9a\right)}^{3/2}-a^{3/2}]$

$=\frac{4}{3}\sqrt{a}[27a^{3/2}-a^{3/2}]=\frac{4}{3}\sqrt{a}\left[26a^{3/2}\right]=\frac{104}{3}{a}^2$