Question

The area between the parabolas $y^2=4ax,x^2=4ay$ is

• A. $\frac{16a^2}{3}$
• B. $\frac{16a^2}{2}$
• C. $\frac{10a^2}{3}$
• D. $\frac{16a^2}{5}$

Answer 1

The correct option is A $\frac{16a^2}{3}$


The given curves are $y^2=4ax,x^2=4ay$
$x^2=4ay\Rightarrow x^4=16a^2{y}^2\Rightarrow x^4=16a^2\left(4ax\right)\Rightarrow x^4=64a^{3}x$
$\Rightarrow x^4-64a^{3}x=0\Rightarrow x(x^3-64a^3)=0\Rightarrow x=0$ or $x=4a$
$x=0\Rightarrow y=0;x=4a\Rightarrow y=4a$
$\therefore$ The two curves intersect at (0, 0), (4a, 4a)
The required region lines between the given curves from x = 0 or x = 4a
The required area = ${\int }_0^{4a}(2\sqrt{a}\sqrt{x}-\frac{x^2}{4a})dx$
$=2\sqrt{a}.\frac{2}{3}{\left[x^{\frac{3}{2}}\right]}_a^{4a}-\frac{1}{4a}{\left[\frac{x^3}{3}\right]}_0^{4a}=\frac{32a^2}{3}-\frac{16a^2}{3}=\frac{16a^2}{3}$sq. unit