The area enclosed by the curve y2 = 4ax and its latus - rectum is

y² = 4ax and latus rectum

area of I = area of II

So,  area = $2 × \int\limits_0^a y d x$

$=2 \times \int\limits_0^a 2 \sqrt{a} \sqrt{x} d x$

$=2 \times 2 \sqrt{a} \times \frac{2}{3}\left[x^{3 / 2}\right]_0^a$

$=\frac{8}{3} a^2$

Option: 1