Explanation

(a)
The equation of latusrectum of the parabola $y^{2} = 12x\ is\ x = 3$
Coordinates of end points of latuserectum are (3,6) and (3,-6)
$Required\ length = 2\int_{0}^{3}{\sqrt{1 + \left( \frac{dy}{dx} \right)^{2}}dx}$
$= 2\int_{0}^{3}{\sqrt{1 + \left( \frac{6}{y} \right)^{2}} = 2\int_{0}^{3}{\sqrt{\frac{12x + 36}{12x}}dx = 2\int_{0}^{3}{\frac{x + 3}{\sqrt{x^{2} + 3x}}dx}}}$
$= 2\left\lbrack \int_{0}^{3}{\frac{2x + 3}{2\sqrt{x^{2} + 3x}}dx + \frac{3}{2}\int_{0}^{3}{\frac{1}{\sqrt{\left( x + \frac{3}{2} \right)^{2} - \left( \frac{3}{2} \right)^{2}}}dx}} \right\rbrack$
$= 2\left\lbrack \sqrt{x^{2} + 3x} + \frac{3}{2}\log\left| \left( x + \frac{3}{2} \right) + \sqrt{x^{2} + 3x} \right| \right\rbrack_{0}^{3}$
$= 2\left\lbrack 3\sqrt{2} + \frac{3}{2}\log{\left( \frac{9}{2} + 3\sqrt{2} \right) - \frac{3}{2}\log\left( \frac{3}{2} \right)} \right\rbrack$
$= 2\left\lbrack 3\sqrt{2} + 3\log\left( 3 + 2\sqrt{2})^{\frac{1}{2}} \right) \right\rbrack\$
$= 2\left\lbrack 3\sqrt{2} + 3\log\left( \sqrt{2} + 1 \right) \right\rbrack$
$= 6\left\lbrack \sqrt{2} + \log\left( 1 + \sqrt{2} \right) \right\rbrack$