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# The area bounded by the parabola $y^{2}=x$ and its latus rectum is

$\begin{array}{1 1}(1)\frac{4}{3}&(2)\frac{1}{6}\\(3)\frac{2}{3}&(4)\frac{8}{3}\end{array}$

The equation of the parabola is $y^2= x$
$4a= 1 => a= \large\frac{1}{4}$
Required area $=2 \int \limits_0^{1/4} y dx$
$\qquad=2 \int \limits _0^{1/4} \sqrt x dx$
$\qquad= 2 \large\frac{x^{1/2+1}}{1/2+1} \bigg]_0^{1/4}$
$\qquad= 2 \times \large\frac{2}{3} \bigg[ \bigg( \large\frac{1}{4}\bigg)^{3/2} -0^{3/2} \bigg]$
$\qquad =\large\frac{4}{3} \bigg [ \frac{1}{(2^2)^{3/2}}-0 \bigg]$
$\qquad =\large\frac{4}{3} \times \frac{1}{8} =\frac{1}{6}$
Hence 2 is the correct answer.