Area bounded by $y^2=4ax$

and latus rectum $=2 \times$ area is first quadrent.

$A= 2 \int \limits_0^a y dx =2 \int \limits _0^a 2 \sqrt a x^{\large\frac{1}{2}} dx$

$\qquad= 4 \sqrt a \bigg [\large\frac{x}{3} $$x^{\large\frac {3}{2}}\bigg]_0^a$

$\qquad= \large\frac{8}{3}$$a^{\large\frac{1}{2}}a^{\large\frac{3}{2}}$

$\qquad=\large\frac{8}{3}$$a^2 sq \;units$