Question

The area (in sq unit)bounded by the parabola $y^2=4ax$ and the line$x=a$ and $x=4a$ is

• A. $\frac{35}{3}{a}^2$
• B. $\frac{4}{3}{a}^2$
• C. $\frac{7}{3}{a}^2$
• D. $\frac{56}{3}{a}^2$

Answer 1

The correct option is D

$\frac{56}{3}{a}^2$

Explanation for Correct Answer:

Determining the area bounded by the curve.

Given, curves are parabola $y^2=4ax$ and the lines $x=a$ and $x=4a$

Now the graph of the given curves is as shown.

Now $y^2=4ax\text{or}y=\sqrt{4ax}$

the area of the bounded region is given as:

Area of the curve $PMNQP$$=2\times$Area of the curve $PSRQP$

Hence,

$$\begin{gathered} A=2{\int }_a^{4a}\sqrt{4ax}dx \\ =4\sqrt{a}{{\left[\frac{x^{{\displaystyle \frac{3}{2}}}}{{\displaystyle \frac{3}{2}}}\right]}_a}^{4a} \\ =\frac{4\times 2}{3}\sqrt{a}\left[{\left(4a\right)}^{\frac{3}{2}}-{\left(a\right)}^{\frac{3}{2}}\right] \\ =\frac{8}{3}\sqrt{a}\times {\left(a\right)}^{\frac{3}{2}}\left[8-1\right] \\ =\frac{8}{3}{a}^2\left[7\right] \\ =\frac{56}{3}{a}^2 \end{gathered}$$

Therefore the area bounded by the given curve is equal to $\frac{56}{3}{a}^2\text{square units}$

Hence, the correct answer is option (D)