Question

The are included between the curves $y^2=4ax$ and $x^2=4ay$ is ____ sq units.

• A. $\frac{16a^2}{3}$
• B. $\frac{8a^2}{3}$
• C. $\frac{4a^2}{3}$
• D. $\frac{5a^2}{3}$

Answer 1

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

We have, $y^2=4ax$ --------------------------- (1)

$x^2=4ay$ ---------------------------- (2)

(1) and (2) intersects hence

$x=\frac{y^2}{4a}$ (a > 0)

$\Rightarrow (\frac{y^2}{4a}{)}^2=4ay$

$\Rightarrow y^4=64a^{3}y$

$\Rightarrow y^4–64a^{3}y=0$

$\Rightarrow y[y^3–(4a{)}^3]=0$

$\Rightarrow y=0,4a$

When y = 0, x = 0 and when y = 4a, x = 4a.

The points of intersection of (1) and (2) are O(0, 0) and A(4a, 4a).

Area$={\int }_b^a[y_1-y_2]dx$

$={\int }_0^{4a}[\sqrt{4ax}-\frac{x^2}{4a}]dx$

${\left[\frac{4\sqrt{a}}{3}\right(x{)}^{\frac{3}{2}}-\frac{x^3}{12a}]}_0^{4a}$

$=\frac{32a^2}{3}-\frac{16a^2}{3}$

$=\frac{16a^2}{3}$