Area bounded by $x=\frac{y^2}{4}$ and its latus rectum is:

$\frac{8}{3}sq.units $

The correct answer is Option (2) → $\frac{8}{3}sq.units $

Assuming y as d

$y^2=4x$

from definition of parabola

$OX'=OX=a$

so $AB=OX'+OX=2a$

By definition of parabola

Now finding area by symmetry 

Area I = Area II

Area = 2 Area I = $2\int\limits_0^a2\sqrt{x}dx$

$=4×\frac{2}{3}\left[x^{3/2}\right]_0^a$

here $a = 1$

$=\frac{8}{3}sq.units $