Let P be a point on the parabola, y ^ 2 = 12x and N be the foot of...

Comparing the equation of the parabola, with standard equation ${y}^{2} = 4ax$, we get $a = 3$.

Figure is drawn on the basis of the given information.

Parametric form of parabola is $x = a{t}^{2}, y = 2at$

$∴ P = \left(3{t}^{2}, 6t\right)$, $N = \left(3{t}^{2}, 0\right)$, $M = \left(3{t}^{2}, 3t\right)$

$⇒ Q = \left(\dfrac{3{t}^{2}}{4}, 3t\right)$

Equation of $NQ$: $y - 0 = \left(\dfrac{3t - 0}{\dfrac{3{t}^{2}}{4} - 3{t}^{2}}\right)\left(x - 3{t}^{2}\right)$

i.e., $y = \dfrac{- 4}{3t} \cdot \left(x - 3{t}^{2}\right) ⇒ y = \dfrac{- 4}{3t} \cdot x + 4t$

$⇒ 4t = \dfrac{4}{3} ⇒ t = \dfrac{1}{3}$

$PN = 6t = 2$

$MQ = 3{t}^{2} - \dfrac{3{t}^{2}}{4} = \dfrac{9{t}^{2}}{4} = \dfrac{1}{4}$