Question

The area (in sq units) of the region bounded by the curves $y^2=4ax$ and $x^2=4ay$, $a>0$ is

• A. $16\left(\frac{a^2}{3}\right)$
• B. $14\left(\frac{a^2}{3}\right)$
• C. $13\left(\frac{a^2}{3}\right)$
• D. $16a^2$
• E. None of these

Answer 1

The correct option is A

$16\left(\frac{a^2}{3}\right)$

Explanation for correct option:

Step 1: Finding the point of intersection

We have the given Equation of the curves as,

$y^2=4ax......\left(1\right)$

$$\begin{gathered} \Rightarrow x^2=4ay \\ \Rightarrow y=\frac{x^2}{4a}......\left(2\right) \end{gathered}$$

Substituting the value of Equation $\left(2\right)$ in Equation $\left(1\right)$ we get

$$\begin{gathered} \Rightarrow {\left(\frac{x^2}{4a}\right)}^2=4ax \\ \Rightarrow \left(\frac{x^4}{16a^2}\right)=4ax \\ \Rightarrow x^4=4ax\times 16a^2 \\ \Rightarrow x^4=64a^{3}x \end{gathered}$$

On further simplification,

$$\begin{gathered} x^4-64a^{3}x=0 \\ \Rightarrow x\left(x^3-64a^3\right)=0 \\ \Rightarrow x=0 \\ \Rightarrow x^3=64a^3 \\ \Rightarrow x=\sqrt[3]{64a^3} \\ \Rightarrow x=4a \end{gathered}$$

These are the points of intersection

The graph is as shown:

Step 2: Finding the area (in sq units) of the region bounded by the curves

Now the required area of the shaded region is:

$$\begin{gathered} A={\int }_0^{4a}\left(\sqrt{4ay}-\frac{x^2}{4a}\right)dx \\ A=2\sqrt{a}x{\left[\frac{x^{{\displaystyle \frac{3}{2}}}}{{\displaystyle \frac{3}{2}}}\right]}_0^{4a}-\frac{1}{4a}{\left[\frac{x^3}{3}\right]}_0^{4a} \\ A=\frac{4}{3}\sqrt{a}\times 8a^2-\frac{16}{3}{a}^2 \\ A=\frac{32}{3}{a}^2-\frac{16}{3}{a}^2 \\ A=a^2\left(\frac{32-16}{3}\right) \\ A=\frac{16}{3}{a}^2\text{sq units} \end{gathered}$$

Therefore, option (A) is the correct answer.